利用數值分析探討應用傑瑞克森不等式上下界之誤差及簡易證明

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: Gerretsen

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Abstract

Due to Gerretsen - inequality always can solve many tough question in proving effectively.

So, this research use method of primary substitution operation and arithmetic average greater

than geometry average to conclude Gerretsen -inequality as below:

16Rr − 5r2 ≤ s2 _{≤ 4R}2_{+ 4Rr + 3r}2_.

And then, using the numerical analysis to conclude the deviation for upper and under limit

of Gerretsen - inequality.

The research found :

1.Gerretsen- inequality only use the character of arithmetic average greater than geometry

average to prove the inequality . The rest is just routine primary substitution operation.

2.Using Excel to calculate upper limit of Gerretsen - inequality, the difference of inequality

inclines to 0.33333.

3.Using Excel to calculate under limit of Gerretsen - inequality, the difference of inequality

inclines to 0.33333.

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1 b2_{+ c}2 + 1 c2_{+ a}2 + 1 a2_{+ b}2 ≤ 1 2bc + 1 2ca + 1 2ab = 1 2( 1 ab+ 1 bc+ 1 ca) = a + b + c 2abc = 2s 2 · 4Rrs = 1 4Rr. (2.20)

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_: a2 h2 b + h2c + b 2 h2 c + h2a + c 2 h2 a+ h2b ≤ R r.

âk

[(a2+ b2) + (b2+ c2) + (c2+ a2)]( 1 a2 _{+ b}2 + 1 b2 _{+ c}2 + 1 c2_{+ a}2) ≥ 9.

1/Rû)ø

1 a2 _{+ b}2 + 1 b2 _{+ c}2 + 1 c2_{+ a}2 ≥ 9 2(a2_{+ b}2_{+ c}2₎.

‚à0

_(2.11),

/â

_^:sR.ø

_: a2+ b2+ c2 ≤ 2[(4R2_{+ 4Rr + 3r}2_{) − 4Rr − r}2_] = 2(4R2 + 2r2).

Ĥ)ƒ

1 a2_{+ b}2 + 1 b2_{+ c}2 + 1 c2_{+ a}2 ≥ 9 4(4R2_{+ 2r}2₎. (2.21)

yâ

_(2.19)(2.21)

ªø

a2 h2 b + h2c + b 2 h2 c + h2a + c 2 h2 a+ h2b ≥ 9R 2 4R2_{+ 2r}2.

]

9R2 4R2_{+ 2r}2 ≤ a2 h2 b + h2c + b 2 h2 c + h2a + c 2 h2 a+ h2b ≤ R r.

¢ÄÑ

9R2 4R2_{+ 2r}2 = 8R2_{+ R}2 4R2_{+ 2r}2 ≥ 8R 2_{+ (2r)}2 4R2_{+ 2r}2 = 2.

ku

_,

û˝6ªJ)ƒ—é.

: a2 h2_b + h2 c + b 2 h2 c + h2a + c 2 h2 a+ h2b = 2.

ù ^:sR.Dúi$qi}(Åí4”

×

_{Ðç6c™âk,ø¹dıFH}

_,

ô

_{.ƒ‚à^:sR.×)7úi$}

qi}(ÅíS4”

,

à-

_: 4_{” 2.3.4.} 1 t2 a + 1 t2 b + 1 t2 c = 2R + r 8Rr2 + 4R + r 8Rr2 .

Ñ„p¤ìÜ

,

.âlõ-ÞíùÜ

: t2_a= 4bcs(s − a) (b + c)2

„p

:

q

_{CD = x, BD = y ,}

âúi$}(4”ø

_{: b : c = x : y}

¢ÄÑ

_{x + y = a}

FJ

x = _b+cab , y = _b+cac

â

_Í·MìÜ

_{(stewart’s theorem)}

ø

AB2· CD + AC2· BD − AD2· BC = BC · BD · CD

¹

c2x + b2y − at2_a= axy t2_a = c 2_{x + b}2_{y − axy} a = c 2_· ab b+c + b 2_· ac b+c+ a · a2_bc (b+c)2 a = abc[c(b + c) 2_{+ b(b + c) − a}2_] a(b + c)2 = bc[(b + c) 2_{− a}2_] (b + c)2 = bc(b + c + a)(b + c − a) (b + c)2 = 4bcs(s − a) (b + c)2

]

t2_a= 4bcs(s − a) (b + c)2

-Þ

,

û˝6ªø¥V„p¤ìÜ

„p

:

ÄÑ

bc t2 a = 1 4 · s s − a + 1 2 + s − a 4s .

FJ

1 t2 a = 1 4bc( s s − a + 1 2+ s − a 4s ) = s 4bc(s − a) +_4bc3 − a2 4abcs .

°Ü

1 t2_b = s 4ca(s − b) + 3 4ca − b2 4abcs . 1 t2 c = s 4ab(s − c) +_4ab3 − c2 4abcs .

¢ÄÑ‚à0

(2.11) (2.12) (2.13)

ªø

1 t2 a + 1 t2 b + 1 t2 c = s 4abc( a s − a + b s − b+ c s − c) + 3 4( 1 ab + 1 bc + 1 ca) − a2_{+ b}2_{+ c}2 4abcs = s abc[s( 1 s − a + 1 s − b+ 1 s − c) − 3] + 3 4( 1 ab + 1 bc + 1 ca) − a2+ b2+ c2 4abcs = s 4 · 4Rrs(s · 4R + r rs − 3) + 3 4 · 1 2Rr − 2(s2_{− 4Rr − r}2₎ 4 · 4Rrs · s = 2R + r 8Rr2 + 4R + r 8Rs2 .

]

1 t2 a + 1 t2 b + 1 t2 c = 2R + r 8Rr2 + 4R + r 8Rs2 .

-Þ

,

Bbú¤4”yTªø¥í«n

Žâ^:sR.ô.Rª)ø.

,

à-

_: 2R + r 8Rr2 + 4R + r 8Rs2 ≤ 2R + r 8Rr2 + 4R + r 8R(16Rr − 5r2₎ ≤ 2R + R 2 8Rr2 + 4R + R₂ 8R(16r · 2r − 5r2₎ = 5 16r2 + 1 48r2 = 1 3r2.

¥v

_,

Zª)ƒ-ÞR

_: 1 t2 a + 1 t2 b + 1 t2 c ≤ 1 3r2.

J/ÑJ

∆ABC

u£úi$v

,

¦U

¢â¤ªø−

(t2_a+ t2_b + t2_c)(1 t2 a + 1 t2 b + 1 t2 c ) ≥ 9.

FJ

t2_a+ t2_b + t2_c ≥ ₁ 9 t2 a + 1 t2 b + _t12 c ≥ 9₁ 3r2 = 27r2.

¥v

_,

û˝6)ƒ.°íR

_:

t2_a+ t2_b + t2_c ≥ 27r2_.

J/ÑJ

_∆ABC

u£úi$v

_,

¦U

ú ^:sR.FRû|5½b.

%¬n

,

û˝6)ƒø_«àÊúi.,íìÜJ£úi$iÅ5Èíø

_.

,

à-

_: 4_{” 2.3.5.}

Ê

_∆ABC

2

_,



tan2A 2 + tan 2B 2 + tan 2C 2 ≥ 2 − 2r R.

Ñ„p¤.

,

lõ-ÞíùÜ

: R(4R + r)2 − (4R − 2r)s2 _{≥ 0.}

„p

:

â«….£

s2 _{≤ 2R}2_{+ 10Rr − r}2_{+ 2(R − 2r)pR(R − 2r)}

øk„

_: R(4R + r)2 − (4R − 2r)s2 ≥ 0,

û

_˝6ÉÛb„p

_: 16R3+ 8R2r + Rr2 ≥ (4R − 2r)[2R2+ 10Rr − r2+ 2(R − 2r)pR(R − 2r)] ⇔ (R − 2r)(8R2_{− 12Rr + r}2₎ ≥ 4(2R − r)(R − 2r)pR(R − 2r) ⇔ (R − 2r)2(8R2− 12Rr + r2)2 ≥ 16R(2R − r)2(R − 2r)3 ⇔ (R − 2r)2_(16R2_r2 _{+ 8Rr}3_{+ r}4_{) ≥ 0.}

,éÍA

_,

Ĥ7-.

R(4R + r)2− (4R − 2r)s2 _{≥ 0.}

ʤ!€,

,

-Þ„p¤ìÜ

„p

:

ÄÑ

1 s − a + 1 s − b+ 1 s − c = 4R + r rs ,

FJ

tanA 2 + tan B 2 + tan C 2 = r s − a + r s − b + r s − c = r( 1 s − a + 1 s − b+ 1 s − c) = r · 4R + r rs = 4R + r s .

1/‚à

tanA 2 tan B 2 + tan B 2 tan C 2 + tan C 2 tan A 2 = 1,

Rû|

tan2 A 2 + tan 2 B 2 + tan 2 C 2 = (tan A 2 + tan B 2 + tan C 2) 2_{− 2(tan}A 2 tan B 2 + tanB 2 tan C 2 + tan C 2 tan A 2) = (4R + r) 2 s2 − 2.

âùÜ

R(4R + r)2− (4R − 2r)s2 _{≥ 0.}

)

(4R + r)2 s2 ≥ 4 − 2r R,

]

tan2 A 2 + tan 2B 2 + tan 2 C 2 ≥ 2 − 2r R.

âk

sec2A 2 = 1 + tan 2 A 2,

ku

_,

û˝6)ƒ

sec2 A 2 + sec 2 B 2 + sec 2 C 2 ≥ 5 − 2r R.

‚à¤4”

_,

û˝6ªJ¡Ë„p-Þííúi.£úi$.

4_{” 2.3.6.}

Ê

_∆ABC

2

_,



cos2A 1 + cos A + cos2B 1 + cos B + cos2C 1 + cos C ≥ 1 2.

„p

:

ÄÑ

cos A + cos B + cos C = 1 + 4 · sinA 2 sin B 2 sin C 2,

/

sinA 2 sin B 2 sin C 2 = r 4R,

FJ

cos A + cos B + cos C = 1 + r

R.

Ĥ

cos2_A 1 + cos A = cos A − 1 + 1 1 + cos A = cos A − 1 + 1 2 cos2 A 2 = cos A − 1 + 1 2sec 2A 2,

âRûø

sec2A 2 + sec 2B 2 + sec 2 C 2 ≥ 5 − 2r R,

FJ

cos2_A 1 + cos A + cos2_B 1 + cos B + cos2_C

1 + cos C = cos A + cos B + cos C − 3

+1 2(sec 2A 2 + sec 2B 2 + sec 2 C 2) ≥ (1 + r R) − 3 + 1 2(5 − 2r R) ≥ 1 2,

]

cos2_A 1 + cos A + cos2_B 1 + cos B + cos2_C 1 + cos C ≥ 1 2. 4_{” 2.3.7.} 3(2R − r) 5R − r ≤ P a2 P ab ≤ 2R2_{+ r}2 (R + r)2.

„p

:

âøí0

(2.10) (2.11)

ªø

a2_{+ b}2_{+ c}2 ab + bc + ca = 2(s2_{− 4Rr − r}2₎ s2_{+ 4Rr + r}2 . = 2 − 4(4Rr + r 2₎ s2_{+ 4Rr + r}2

y

_{Žâ^:sR.ø}

4Rr + r2 (R + r)2 ≤ 4(4Rr + r2) s2_{+ 4Rr + r}2 ≤ 4R + r 5R − r.

1/

2 −4R + r 5R − r ≤ 2 − 4(4Rr + r2₎ s2_{+ 4Rr + r}2 ≤ 2 −4Rr + r 2 (R + r)2

¹

2 − 4R + r 5R − r ≤ P a2 ab + bc + ca ≤ 2 − 4Rr + r 2 (R + r)2.

)

3(2R − r) 5R − r ≤ a2_{+ b}2 _{+ c}2 ab + bc + ca ≤ 2R 2_{+ r}2 (R + r)2. (2.22)

M)øTíu

(2.22)

.õÒ,Hb.

a2_{+ b}2 _{+ c}2 _{≥ ab + bc + ca}

_Êú

i$2í‹#

Í7

,

¤

_{¹dıT6 wž FTƒ^:sR.Ê„pÉúi$íS.}

2@à

_}˜

_,

Ä

_(2.8)

$

Ý

,

7/˝¬si.·'#íŠ?

,

ʤ

7^:sR.íø_‹#

4_{” 2.3.8.} C + 16Rr − 5r2 ≤ s2 _{≤ 4R}2_{+ 4Rr + 3r}2_{− C. (2.23)}

w2

_{C =} 1

2R|(a − b)(b − c)(c − a)|,

1/ç

∆ABC

Ñ£úi$v

,

Uÿ}A

„p

:

ç

_(2.23)

2˝i.gk

(a − b)2(b − c)2(c − a)2 ≤ 4R2_(s2_{− 16Rr + 5r}2₎2_{. (2.24)}

«àúi$0

ªJ)ƒ

_(2.24)

2

−4r2_(s2_{− 2R}2_{− 10Rr + r}2₎2_{+ 16Rr}2_{(R − 2r)}3 _{≤ 4R}2_(s2_{− 16Rr + 5r}2₎2_. ⇔ −4r2_(s2_{− 2R}2_{− 10Rr + r}2₎2_{+ 16Rr}2_{(R − 2r)}3 ≤ 4R2[s2 − 2R − 10Rr + r2+ 2(R − 2r)(R − r)]2. ⇔ (R2+ r2)(s2 − 2R2_{− 10Rr + r}2_{) + 4R}2_{(R − 2r)(R − r)(s}2_{− 2R}2 −10Rr + r2_{) + 4R(R − 2r)}2_(R3_{− 2R}2_{r + 2r}3_{) ≥ 0.} ⇔ (R2+ r2)[s2− 2R2− 10Rr + r2+ +2R 2_{(R − 2r)(R − r)} R2_{+ r}2 ] 2 +8r 5_{R(R − 2r)}2 R2_{+ r}2 ≥ 0. (2.26)

(2.26)

éÍA

,

FJ

(2.24)

A

;(2.23)

2˝i.)„

Í7

, (2.23)

2¬i.gk

(a − b)2(b − c)2(c − a)2 ≤ 4R2(4R2+ 4Rr + 3r2− s2)2. (2.27)

«àúi$0

(2.25)

)

_(2.27)



−4r2_(s2_{− 2R}2_{− 10Rr + r}2₎2_{+ 16Rr}2_{(R − 2r)}3 _{≤ 4R}2_(4R2_{+ 4Rr + 3r}2_{− s}2₎2_. ⇔ −4r2(s2− 2R2− 10Rr + r2)2+ 16Rr2(R − 2r)3 ≤ 4R2_[−(s2_{− 2R}2_{− 10Rr + r}2_{) + 2(R − 2r)(R − r)]}2_. ⇔ (R2_{+ r}2_)(s2_{− 2R}2 _{− 10Rr + r}2₎2_{− 4R}2_{(R − 2r)(R − r)(s}2_{− 2R}2 _{− 10Rr + r}2₎ +4R(R − 2r)2(R3− 2R2r + 2r3) ≤ 0. ⇔ (R2_{+ r}2_)[s2_{− 2R}2 _{− 10Rr + r}2₋ 2R2(R − 2r)(R − r) R2_{+ r}2 ] 2 +8r 5_{R(R − 2r)}2 R2_{+ r}2 ≥ 0. (2.28)

(2.28)

éÍA

,

FJ

(2.27)

A

; (2.23)

2¬i.)„
J„p¬˙

ªø

_(2.23)

2

,

ç

_∆ABC

Ñ£úi$v

_,

Uÿ}A
]¤ìÜ)„ç

_∆ABC

«

P a ra ≥ 2 √ 3

í‹#

,

‚à×Ðç6Ùždı2

,

w2

_,

 -ø_

S.

: a ra + b rb + c rc ≥ 2√3. (2.29)

ø

_(2.29)

‹#Ñ

a ra + b rb + c rc ≥ √ 2(4R + r) 4R2_{+ 4Rr + 3r}2. (2.30)

„p

_:

q

_∆ABC

íÞ  š¶Å}Ñ

_∆



_s,

1/â

_ra= ∆ (s−a)

í

,

ªø

(2.30)



gk

1 ∆[a(s − a) + b(s − b) + c(s − c)] ≥ 2(4R + r) √ 4R2_{+ 4Rr + 3r}2.

¹

1 ∆[s · (a + b + c) − (a 2 + b2+ c2)] ≥ √ 2(4R + r) 4R2_{+ 4Rr + 3r}2. (2.31)

«àƒ0

(2.11)

¸

∆ = rs

)ø

_(2.31)

gk

2(4R + r) s ≥ 2(4R + r) √ 4R2_{+ 4Rr + 3r}2.

¹

s2 ≤ 4R2_{+ 4Rr + 3r}2_{. (2.32)}

7

(2.32)

u

_O±í^:sR.

_,

Ä7

(2.30)

A

-Þ„p

(2.30)

ª

_(2.29)

#

_,

éÍ7ø

,

Û„p-.

4R + r √ 4R2_{+ 4Rr + 3r}2 ≥ √ 3. (2.33)

ø

_(2.33)

sij

_,

cÜ)

(R − 2r)(R + r) ≥ 0. (2.34)

yâ«….ø

(2.34)

A

,

Ä7

(2.33)

A

,

]ªzp

(2.30)

ª

_(2.29)

#

4_{” 2.3.9.} 4 − 2r R ≤ ha ra +hb rb + hc rc ≤ 2R r + 2r R − 2.

„p

:

âúi0

(2.9) (2.10) (2.19)

)ø

ha ra +hb rb +hc rc = 2∆ a ∆ s−a + 2∆ b ∆ s−b + 2∆ c ∆ s−c = 2(s − a a + s − b b + s − c c ) = 2s(1 a + 1 b + 1 c) − 6 = 2s ab + bc + ca abc − 6 = 2s ·s 2_{+ 4Rr + r}2 4Rrs − 6 = 1 2Rr(s 2_{+ 4Rr + r}2_{) − 6.}

yâ

_^:sR.ª)

1 2Rr(16Rr − 5r 2_{+ 4Rr + r}2_{) − 6 ≤} ha ra + hb rb +hc rc ≤ 1 2Rr(4R 2_{+ 4Rr + 3r}2_{+ 4Rr + r}2_{) − 6,}

¹

4 − 2r R ≤ ha ra +hb rb + hc rc ≤ 2R r + 2r R − 2.

]¤.)„

4_{” 2.3.10.} 4(R − r r ) ≤ bc r2 a + ac r2 b + ab r2 c ≤ 4(R − r r ) 2_.

J/ÑJ

_∆ABC

Ñ£úi$v

_,

UA

‚àúiç2ít

_,

Bb

a = r cos A 2 sinB 2 sin C 2 , b = r cos B 2 sinA₂ sinC₂ , c = r cos C 2 sinA₂ sinB₂ .

£

ra = r cot B 2 cot C 2, rb = r cot A 2 cot C 2, rc = r cot A 2 cot B 2.

Hp

M = bc r2 a +ac r2 b +ab r2 c ,

)

M = cot2A 2 tan B 2 tan C 2 + cot 2B 2 tan A 2 tan C 2 + cot 2C 2 tan A 2 tan B 2 + tanA 2 tan B 2 + tan B 2 tan C 2 + tan C 2 tan A 2 = cot 3 A 2 + cot 3 B 2 + cot 3 C 2

cotA₂ cotB₂ cotC₂ + 1.

‚à0

x3+ y3 + z3− 3xyz = (x + y + z)(x2_{+ y}2 _{+ z}2_{− xy − yz − zx).}

1·<ƒ

cotA 2 + cot B 2 + cot C 2 = cot A 2cot B 2cot C 2,



M = cot2A 2 + cot 2B 2 + cot 2C 2 − cot A 2 cot B 2 − cot B 2 cot C 2 − cot C 2 cot A 2 +3 + 1 = (cotA 2 + cot B 2 + cot C 2) 2_{− 3(cot}A 2 cot B 2 + cot B 2 cot C 2 + cot C 2 cot A 2) +4.

y·<ƒ

cotA 2 + cot B 2 + cot C 2 = s r.

J£

cotA 2 cot B 2 + cot B 2 cot C 2 + cot C 2 cot A 2 = 4R + r r .

†

M = s 2 r2 − 3 4R + r r + 4 = s 2_{− 12Rr − 3r}2_{+ 4r}2 r2 .

yâ

_^:sR.ª)ø

_,

à-4Rr − 4r2 r2 ≤ M ≤ 4R2_{− 8Rr + 4r}2 r2 ,

¹

4(R − r r ) ≤ M ≤ 4( R − r r ) 2_.

Ĥ)„

4_{” 2.3.11.} ra ra− 2r + rb rb− 2r + rc rc − 2r ≤ 9;

J/ÑJ

_∆ABC

Ñ£úi$v

_,

UA

„p

:

ÄÑ

ra ra− 2r + rb rb− 2r + rc rc− 2r = P ra(rb− 2r)(rc− 2r) Q(ra− 2r) = 3Q ra− 4(P rarb)r + 4(P ra)r 2 Q ra− 2(P rarb)r + 4(P ra)r2− 8r3 ,

¢âúi0

: (A) ra· rb· rc = rs2. (B) rarb+ rbrc + rcra = s2. (C) ra+ rb+ rc = 4R + r.

FJ

ra ra− 2r + rb rb− 2r + rc rc− 2r = s 2_{− 16Rr − 4r}2 s2_{− 16Rr + 4r}2 ≤ 9.

yâ

_^:sR.

_,

ª)ø¤ìÜA

4_{” 2.3.12.} ra+ r ra− r +rb+ r rb− r + rc+ r rc− r ≥ 9 − 3r 2R.

J/ÑJ

∆ABC

Ñ£úi$v

,

UA

„p

_:

ÄÑ

ra+ r ra− r + rb + r rb− r +rc+ r rc− r = P(ra+ r)(rb + r)(rc+ r) Q(ra− r) = 3Q ra− (P rarb)r + (P ra)r 2_{+ 3r}3 Q ra− (P rarb)r + (P ra)r2 − r3 .

yâ,HìÜ2í0

,

ª)ø

ra+ r ra− r + rb+ r rb − r +rc+ r rc− r = s 2_{+ 2Rr + 2r}2 2Rr ≥ 16Rr − 5r 2_{+ 2Rr + 2r}2 2Rr = 9 − 3r 2R.

]¤.

_¹ª)„

û ‚à^:sR.«ú_à.í-ä,l

;W„.õ`¤)Oíà.2

_,

FY“7-Þú_.

sqPÀ.

: b2_{+ c}2_{− a}2 bc + a2_{+ b}2_{− c}2 ab + c2_{+ a}2_{− b}2 ca ≤ 3 ≤ ( a b) 2_{+ (}b c) 2_{+ (}c a) 2_.

é,R

(Anderson)

.

: a3(s − a) + b3(s − b) + c3(s − c) ≤ s · abc.

l.gs

(Morghescu)

.

b3_{+ c}3 _{− a}3 b + c − a + a3_{+ b}3_{− c}3 a + b − c + c3_{+ a}3_{− b}3 c + a − b ≤ ab + bc + ca.

%¬«n

,

_{°6Û˛)ƒç6í-ä,l}

4_{” 2.3.13.} Xb2 + c2− a2 bc ≥ 7 − 2R r .

„p

:

ÄÑ

b2+ c2 ≥ 2bc, abc = 4Rrs. b2_{+ c}2_{− a}2 bc ≥ 2bc − a2 bc = 2 − a3 abc = 2 − a3 4Rrs.

/

a3+ b3+ c3 = 2s(s2− 6Rr − 3r2_).

FJ

b2_{+ c}2_{− a}2 bc + a2_{+ b}2_{− c}2 ab + c2_{+ a}2_{− b}2 ca ≥ (2 − a3 4Rrs) + (2 − b3 4Rrs) + (2 − c3 4Rrs) = 6 − a 3_{+ b}3_{+ c}3 4Rrs = 6 − 2s(s2− 6Rr − 3r2₎ 4Rrs = 6 − s 2_{− 6Rr − 3r}2 2Rr .

â

_^:sR.ø

b2_{+ c}2_{− a}2 bc + a2_{+ b}2_{− c}2 ab + c2_{+ a}2_{− b}2 ca ≥ 6 − (4R2_{+ 4Rr + 3r}2_{) − 6Rr − 3r}2 2Rr = 6 − 4R 2_{− 2Rr} 2Rr = 7 − 2R r .

]

b2_{+ c}2 _{− a}2 bc + a2_{+ b}2 _{− c}2 ab + c2_{+ a}2_{− b}2 ca ≥ 7 − 2R r .

âìýìÜ

_a2 _{= b}2_{+ c}2_{− 2bc cos A}

_ª)

b2_{+ c}2_{− a}2 bc = 2 cos A

!¯sqPÀ.

7 −2R

r ≤ 2 cos A + 2 cos B + 2 cos C ≤ 3

ku

_,

Zª)ƒø_gíúi.

,

R|

7

2−

R

r ≤ cos A + cos B + cos C ≤

3 2.

4_{” 2.3.14.}

a3(s − a) + b3(s − b) + c3(s − c) ≥ (8r

„p

_:

ÄÑ

a4+ b4+ c4 = 2(s2− 4Rr − r2₎2_{− 8r}2_s2_{, s ≤} 4R + r_√ 3 ,

FJ

a4+ b4+ c4 = 2[s2− r(4R + r)]2_{− 8r}2_s2 _{≤ 2(s}2₋√_3rs)2_{− 8r}2_s2 = 2s2(s −√3r)2− 8r2s2.

1/

a3_{(s − a) + b}3_{(s − b) + c}3_{(s − c)} s · abc = s(a3_{+ b}3_{+ c}3_{) − (a}4_{+ b}4 _{+ c}4₎ s · abc ≥ 2s 2_(s2_{− 6Rr − 3r}2_{) − 2s}2_{(s −}√_3r)2_{+ 8r}2_s2 s · 4Rrs = 4 √ 3rs − 12Rr − 4r2 4Rr = √ 3rs − 3Rr − r2 Rr = √ 3s − 3R − r R .

¢ÄÑq„|

s ≥ 3√3r,

FJ

a3_{(s − a) + b}3_{(s − b) + c}3_{(s − c)} s · abc ≥ 9r − 2R − r R = 8r R − 2.

]

a3(s − a) + b3(s − b) + c3(s − c) ≥ (8r R − 2)s · abc. 4_{” 2.3.15.} b3_{+ c}3 _{− a}3 b + c − a + a3_{+ b}3 _{− c}3 a + b − c + c3_{+ a}3_{− b}3 c + a − b ≥ (4 − 3R 2r)(ab + bc + ca).

„p

_:

*úi$0

(2.10) (2.11)

)ø

2(ab + bc + ca) − (a2+ b2+ c2) = 2(s2+ 4Rr + r2) − 2(s2− 4Rr − r2₎ = 16Rr + 4r2 = 4r(4R + r). (2.35)

¢ÄÑ

1 s − a + 1 s − b+ 1 s − c = 4R + r rs ,

FJ

Xb3+ c3− a3 b + c − a = X[(b + c)3− a3] − 3bc(b + c − a) − 3abc b + c − a = X[(b + c)2+ (b + c)a + a2− 3bc] −X 3abc b + c − a = 3Xa2+Xbc − 3X 4Rrs 2(s − a) = 3Xa2+Xbc − 6RrsX 1 s − a = 3Xa2+Xbc − 6Rrs4R + r rs = 3Xa2+Xbc − 6R(4R + r) = 3Xa2+Xbc − 3R 2r · 4r(4R + r). (2.36)

â

_(2.35)(2.36)

)

Xb3+ c3− a3 b + c − a = 3 X a2+Xbc − 3R 2r · (2 X bc −Xa2) = (3 + 3R 2r) X a2+ (1 − 3R r ) X bc,

Í7«à-Þ.

, (a2+ b2+ c2) − (ab + bc + ca) = 1 2[(b − c) 2_{+ (a − b)}2_{+ (c − a)}2_] ≥ 0,

Ĥªø

a2+ b2+ c2 ≥ ab + bc + ca,

/

Xb3+ c3− a3 b + c − a ≥ (3 + 3R 2r) X bc + (1 − 3R r ) X bc.

]

Xb3+ c3 − a3 b + c − a ≥ (4 − 3R 2r) X bc.

ü ^:sR.Dúi$}( í.û˝ãH

*×Ðç6"j#|à-.

: 2∆2 R ≤ tatbtc ≤ ∆2 r . (2.37)

d

_[2]

#|,˝«í‹#

_: tatbtc ≥ 27 2 Rr 2 . (2.38)

,H½æù–7.=ß6íÉ·

_;

ã¯Ì#|-‹#$

_: 14(R − r)r2 ≤ tatbtc ≤ (16R − 5r)r2. (2.39)

Í7

,

_{ÇøPç6#|7à-íø_‹#}

: tatbtc ≤ 8Rrs2 9R − 2r. (2.40)

-H.

: tatbtc ≤ 16R2r(cos A 2 cos B 2 cos C 2) 2 _≤ 27 4 R 2_{r. (2.41)}

O

_(2.41)

2í

16R2r(cosA 2 cos B 2 cos C 2) 2_,

¹

(2.37)

í¬«

∆_r2,

¥uÄÑ

: s2_r2 r = 16R 2_r(cosA 2 cos B 2 cos C 2) 2_. ⇔ s = 4R cosA 2 cos B 2 cos C 2.

7âøíúi0

,

sin A + sin B + sin C = 4 cosA 2 cos B 2 cos C 2.

J££ýìÜ

,

)

_: 4R cosA 2 cos B 2 cos C

2 = R(sin A + sin B + sin C)

= 1 2(a + b + c) = s.

yâ×Ðç6Ï?_F#.6q„u

(2.37)

˝«í‹#

: tatbtc ≥ 3 √ 3r∆. (2.42)

|¡

,

×Ðç6óÁ‰°v6#7

(2.37)(2.38)

íy‹#

128 9 Rr 2₋ 13 9 r 3 _{≤ t} atbtc ≤ (16R − 5r)2. (2.43)

J,®J/ÑJ

_∆ABC

Ñ£úi$v

_,

¦U

O×Ðç6óÁ‰í„pœõ

,

/(ø¶}F„

(2.42)

#k

_(2.37)

Ï

Ѥ

_,

…dl#|

(2.43)

œ¡í„p

„p

:

ÄÑ

ta = 2bc cosA 2 b + c = 2∆ (b + c) sinA₂ = 2sr (2s − a) sinA₂.

°Ü

_, 2sr (2s − b) sinB₂ , 2sr (2s − c) sinC₂.

¢

sinA 2 sin B 2 sin C 2 = r 4R,

FJ

tatbtc = 32Rr2_s2 (2s − a)(2s − b)(2s − c).

yâ0

(2.9) (2.10) (2.14)

ª)

_: tatbtc = 16Rr2_s2 s2 _{+ 2Rr + r}2.

¥š

,

k„

(2.43)

˝«A

16Rr2_s2 s2_{+ 2Rr + r}2 ≥ 128 9 Rr 2₋ 13 9 r 3_. ⇐⇒ (16R + 13r)s2 ≥ r(128R − 13r)(2R + r).

¥â^:sR.¸«….

,

_¹ª„)

úk

_(2.43)

¬«

,

_ÉÛ„

16Rr2_s2 s2_{+ 2Rr + r}2 ≤ 16Rr 2_{− 5r}2_. ⇐⇒ 5s2 ≤ (2R + r)(16R − 5r).

yâ

_^:sR.

_,

ªø

_{−ÉÛb„p.}

_,

à-

_: 5(4R2+ 4Rr + 3r2) ≤ 32R2+ 6Rr − 5r2. ⇐⇒ (R − 2r)(6R + 5r) ≥ 0.

Í7

_,

â«….

,

_¹ªøA

ã¯J,®

,

Bb.#|Ý7í.

,

Ĥ6ÿzp7

(2.43)

#k

_(2.37): (A) 2∆_R2 ≤ 3√3r∆ ≤ 27 2 Rr 2 _{≤ (14R − r)r}2 _≤ 128Rr2− 13r2 9 ≤ tatbtc ≤ (16R − 5r)r2 ≤ ∆2 r ≤ 27 4 R 2_r. (B) 2∆_R2 ≤ 3√3r∆ ≤ 27 2 Rr 2 _{≤ (14R − r)r}2 _≤ 128Rr 2_{− 13r}2 9 ≤ tatbtc ≤ 8R 9R − 2r · ∆2 r ≤ ∆2 r ≤ 27 4 R 2_r.

-

_ÞÉÛ„p

_(a)

ƒ

_(b)

®A

,

I„à-

:

(a)(b)(f ) ⇐⇒ ∆ ≤ 3 √ 3 2 Rr.

¢

∆ = sr ⇐⇒ s ≤ 3 √ 3 2 R. ⇐⇒ a + b + c ≤ 3√3R.

¤â×Ðç6ïŠ)øA

*

_{(c) (d) (h)}

ªâ«….q„
y*

(e)

ªâ

_^:sR.R)

,

1

/

_(g)

ªâ

Pecaric

¸

_Voloner

í-R)

: (a + b)(b + c)(c + a) ≥ 4sr(9R − 2r)

FJ

tatbtc ≤ 32sR∆2 4sr(9 − 2r) = 8R 9R − 2r · ∆2 r ,

¤

_¹

_(2.40)



ý ‚à^:sR.Rû

P 1 a2

í,-ä

,l

×

_Ðç6Ø£dT|

P 1 a2

í-ä

,l

,

íl

,

×Ðç6éPdı2

,

„p7ú

Ý@iúi$í.

1 a2 + 1 b2 + 1 c2 ≥ 5 4s. (2.44)

J/ÑJ

_∆ABC

Ñ8úi$v

, (2.44)

2íUA

4_{” 2.3.16.}

úkL<úi$Bb

1 a2 + 1 b2 + 1 c2 ≥ 5 9Rr − 1 9R2. (2.45)

„p

_:

â

(ab + bc + ca)2 = a2b2+ b2c2+ c2a2+ 2abc(a + b + c).

£

(A) ab + bc + ca = s2+ 4Rr + r2. (B) abc = 4Rrs.

Žâ0ª)

1 a2 + 1 b2 + 1 c2 = s4+ (2r2− 8Rr)s2_{+ (4Rr + r}2₎2 16R2_r2_s2 .

.

(2.45)

gk

H(s2) = s4+2 9(17r 2 _{− 76Rr)s}2_{+ (4Rr + r}2_{) ≥ 0. (2.46)}

Í7

1 9(76Rr − 17r 2 ) < 16Rr − 5r2 ⇐⇒ R > 7 17r.

ĤäéÍA

â

_{ùŸƒbíÇd£^:sR.}

_,

b„.

(2.46)

_ÉÛ„p

,

à-

_: H(16Rr − 5r2) ≥ 0.

/

H(16Rr − 5r2) = 16 9 r 2_(R2_{− 4Rr + 4r}2₎ = 16 9 r 2_{(R − 2r)}2 _{≥ 0.}

FJ¤.

_(2.45)

)„

Í7

,

Ék

P 1 a2

,ä

,l

, 4_{” 2.3.17.} 1 a2 + 1 b2 + 1 c2 ≤ (R2+ r2)2+ Rr(2R − 3r)2 R2_r3_{(16R − 5r)} .

J/ÑJ

∆ABC

Ñ£úi$v

,

†UA

,

Ñ7„p·æ

,

l„à-íùÜ

ùÜ

: 1 a + 1 b + 1 c ≤ (R + r)2 R∆ .

J/ÑJ

∆ABC

Ñ£úi$v

,

†UA

„p

_:

âøí0

_{(2.9) (2.10)}

)ø-.

1 a + 1 b + 1 c ≤ (R + r)2 R∆ . ⇔ ab + bc + ca abc ≤ (R + r)2 R∆ . ⇔ ab + bc + ca 4R∆ ≤ (R + r)2 R∆ . ab + bc + ca ≤ 4(R + r)2. ⇔ s2+ 4Rr + r2 ≤ 4R2+ 8Rr + 4r2. ⇔ s2 _{≤ 4R}2_{+ 4Rr + 3r}2_.

¥uBbøí^:sR.

,

]ùÜ)„

1/Ê„pJ-íìÜ

„p

:

†

1 a2 + 1 b2 + 1 c2 = a2_b2_{+ b}2_c2_{+ c}2_a2 a2_b2_c2 = (ab + bc + ca abc ) 2₋ 2(a + b + c) abc = (1 a + 1 b + 1 c) 2 ₋2 · 2s 4R∆.

âùÜ)

1 a2 + 1 b2 + 1 c2 ≤ (R + r)4 R2_∆2 − 4s 4R · sr = (R + r) 4 R2_{· s}2_r2 − 1 Rr = (R + r) 4 R2_r2 · 1 s2 − 1 Rr.

â

_^:sR.ª)

1 a2 + 1 b2 + 1 c2 ≤ (R + r)4 R2_r2 · 1 r(16R − 5r) − 1 Rr = (R + r) 4_{− Rr}2_{(16R − 5r)} R2_r3_{(16R − 5r)} = R 4 _{+ 4R}3_{r − 10R}2_r2 _{+ 9Rr}3_{+ r}4 R2_r3_{(16R − 5r)} = (R 2_{+ r}2₎2_{+ 4R}3_{r − 12R}2_r2_{+ 9Rr}3 R2_r3_{(16R − 5r)} = (R 2_{+ r}2₎2_{+ Rr(2R − 3r)}2 R2_r3_{(16R − 5r)} .

y

_{ã¯ùܸ^:sR.ø}

_,

J/ÑJ

∆ABC

Ñ£úi$v

,

¤U}A

,

ĤìÜ)„

þ ^:sR.D¶ä2õúi$í4”

q

∆ABC

úiÑ

a,b,c

/¶ä2õúi$ó@iÑ

a0,b0,c0,

†

a0 a + b0 b + c0 c ≥ 3 2.

„p

:

q

D,E,F

}Ñ

_∆ABC

í

_BC,CA,AB

i,í¶ä2õ

,

*

_E,F

T

_{EM ⊥BC}

k

_{M ,F N ⊥BC}

k

_{N (}

à¬Ç

_),

†

a0 = EF ≥ M N

= a − (BF cos B + CE cos C).

Ĥ,

_¹Ñ

_a0 _{≥ a − (s − a)(cos B + cos C).}

yâÉk

b0,c0

íéN.

,

úó‹

,

)

a0 a + b0 b + c0 c ≥ 3 + (2 − s a − s b − s

c)(cosA + cosB + cosC)

+s(cos A a + cos B b + cos C c ),

ø

cos A + cos B + cos C = 1 + r

R, s a + s b + s c = s2_{+ 4Rr + r}2 4Rr ,

cot A + cot B + cot C = s

2_{− 4Rr − r}2 4sr ,

Hp,˝i

,

kU

3 + (2 − s 2 _{+ 4Rr + r}2 4Rr )(1 + R r) + R 2R( s2− 4Rr − r2 2sr ) ≥ 3 2,

õu´ª?“

: 3 2 − ( s2+ 4Rr + r2 4Rr )(1 + R r) + ( s2− 4Rr − r2 4sr ) ≥ 0,

¹

s2 _{≤ 6r}2_{+ 2Rr − r}2_;

Oâ

_{^:sR.ÉÛ„}

4R2+ 4Rr + 3r2 ≤ 6R2_{+ 2Rr − r}2_,

¹

2R2− 2Rr − 4r2 _{≥ 0;}

â«….ø

, 2R2− 2Rr − 4r2 _{= 2(R − 2r)(R − r) ≥ 0}

ÿ ×ÐÝjæO«

₍₆₉₎

æ

X r2_a h2 b + h2c ≥X r 2 a h2 a+ ra2 (2.47)

„pvúi$.u´A

_?

„p

_:

‚à×Ðç6„d˘ídıªø−

2 − 2 · r 2 R2 ≥ X r2_a h2 a+ ra2 ≥ 7R + r 5R (2.48)

â

_(2.48)

ø

_,

b„

_(2.47)

A

_,

_ÉÛ„

X r_a2 h2 b + h2c ≥ 2 − 2 · r 2 R2 (2.49)

â5a.

X r2_a h2 b + h2c ≥ (P ra) 2 2P h2 a (2.50)

â

_(2.50)

ø

_,

b„

(2.49)

A

,

_ÉÛ„

(P ra)2 2P h2 a ≥ 2 − 2 · r 2 R2 (2.51)

â

_ra= ∆ s−a, ha= 2∆ a

£cúi$0ªø

X ra = X ∆ s − a = sr ·X 1 s − a = sr ·P(s − b)(s − c) (s − a)(s − b)(s − c) = sr · r(4R + r) r2_s = 4R + r. (2.52)

X h2_a = X4∆ 2 a2 = 4s 2_r2_{[(P bc)}2_{− 2abcP a]} (abc)2 = 4s 2_r2_[(s2_{+ 4Rr + r}2₎2 _{− 2 · 4sRr · 2s]} (4sRr)2 = (s 2_{+ 4Rr + r}2₎2_{− 16s}2_Rr 4R2 . (2.53)

â

_{(2.52) (2.53)}

ø

_,

b„

(2.51)

A

,

_ÉÛ„

2R2(4R + r)2 (s2 _{+ 4Rr + r}2₎2_{− 16s}2_Rr ≥ 2 − 2 · r2 R2. (2.54) ⇔ R4(4R + r)2 ≥ [(s2+ 4Rr + r2)2− 16s2Rr](R2− r2) ⇔ f (s2_{) = (R}2_{− r}2_)s4_{− (8R}3_{r − 2R}2_r2_{− 8Rr}3_{+ 2r}4_)s2_{− 16R}6_{− 8R}5_r +15R4r2+ 8R3r3− 15R2_r4_{− 8Rr}5_{− r}6 _{≤ 0. (2.55)}

âk

_{f (s}2₎

_uÉk

_s2

_í

_ùŸƒb

_,

_/Ǩ²,

_,

_b„

_{f (s}2_{) ≤ 0,}

_â

_^:sR.

)ø

_ÉÛb„p

f (16Rr − 5r2) ≤ 0 (2.56)

/

f (4R2+ 4Rr + 3r2) ≤ 0 (2.57)

¹ª)„

íl

,

lû|

_(2.56)

í.

f (16Rr − 5r2) = −16R6− 8R5_{r + 143R}4_r2_{− 80R}3_r3_{− 128R}2_r4 _{+ 80Rr}5_{− 16r}6 = −(R − 2r)(16R5+ 40R4r − 63R3r2− 46R2r3+ 36Rr4 − 8r5) = −(R − 2r)[(16R4+ 72R3r + 81R2r2+ 116Rr3+ 268r4)(R − 2r) +528r5].

Ó(

_,

yRû

_(2.57)

í.

f (4R2+ 4Rr + 3r2) = −8R5r + 15R4r2+ 16R3r3 − 16R2_r4_{− 16Rr}5_{− 16r}6 = −r(R − 2r)(8R4+ R3r − 14R2r2− 12Rr3_{− 8r}4₎ = −r(R − 2r)[(8R3+ 17R2r + 20Rr2+ 28r3)(R − 2r) + 48r4].

Žâ«….¹ª„p|

(2.56)

¸

_(2.57)

í.
]

(2.55)

A

,

*7)

ø

_{(2.54) (2.51) (2.49)}

A

,

⤪ø

(2.47)

)„



_{^:sR.í‹#}

ÊÇán‹#5‡

_,

û˝6.âl!…í–1

_,

à-ú_Ć

_: Lemma 2.3.1. 2R2_{+ 10Rr − r}2_{− C ≤ s}2 _{≤ 2R}2_{+ 10Rr − r}2_{+ C.}

w2

C = r2− 2(R − 2r)√R2_{− Rr. (2.58)} Lemma 2.3.2.

ç

−1 ≤ x ≤ 1

¸

_{0 < α < 1}

v

_,

û˝6øª)ƒ

(1 + x)α = 1 +Xα(α − 1)(α − 2) · · · (α − n + 1) n! · x n_{. (2.59)}

J£

(1 + x)α ≤ 1 + αx. (2.60) Lemma 2.3.3.

cq

−1 < x < 1,

Žâ

(2.59)

ªø

1 1 − x = X xn. (2.61)

‚à,Þ

_Lemma

!…–1ªJß‚)ø-ìÜ

4_{” 2.3.18.} 16Rr − 5r2+ C1 ≤ s2 ≤ 4R2+ 4Rr + 3r2− C1. (2.62)

w2

C1 = r2_{(R − 2r)} R − r ,

„p

:

;W

16Rr − 5r2+ C2 ≤ s2 ≤ 16Rr − 5r2− C2. (2.63)

w2

C2 = 2(R − 2r)(R − r − √ R2_{− 2Rr),}

Í7

,

‚à«….J£

_+›‰

(Bernoulli)

.

1 1−x =P x n_,

_û˝6ø)ƒ

R − r ≥ 0, 0 < r R − r ≤ 1

¸

R − r −√R2_{− 2Rr = (R − r)[1 −} s R2_{− 2Rr} (R − r)2 ] = (R − r)[1 − s 1 − r 2 (R − r)2] ≥ 1 2(R − r)( r R − r) 2 = r 2 2(R − r).

â

_(2.62)

ÿªÄ¤

_(2.62)

)„

4_{” 2.3.19.} 16Rr − 5r2+ C3 ≤ s2 ≤ 4R2+ 4Rr + 3r2− C3. (2.64)

w2

C3 = r(R − 2r) ∞ X n=1 (2n − 3)!! 2n−1_n! ( r R − r) 2n−1_,

„p

_:

‚à

_(2.62)

BbªJ×)-

R − r −√R2_{− 2Rr = (R − r)[1 −} s 1 − r 2 (R − r)2],

íl

r R − r = x(0 < x ≤ 1),

ç

R − r −√R2 _{− 2Rr = (R − r)(1 −}√_{1 − x}2_).

1/*

_(2.59)

û˝6)ø

√ 1 − x2 _{= 1 −} 1 2x 2 ₋ ∞ X n=2 (2n − 3)!! 2n_n! · x 2n_{. (0 < x ≤ 1)}

C

1 −√1 − x2 ₌ 1 2x[x + ∞ X n=2 (2n − 3)!! 2n−1_n! x 2n−1_] = r 2(R − r) ∞ X n=1 (2n − 3)!! 2n−1_n! ( r R − r) 2n−1 .

5(Ó,HªRû|

R − r −√R2_{− 2Rr =} 1 2r ∞ X n=1 (2n − 3)!! 2n−1_n! ( r R − r) 2n−1_{. (2.65)}

FJ*

(2.63)

¸

_{(2.65) ,}

ï,ª|

(2.64),

Ĥ

_(2.64)

)„

4_{” 2.3.20.} 16Rr − 5r2+ r(R − 2r)ς ≤ s2 ≤ 4R2_{+ 4Rr + 3r}2_{− r(R − 2r)ς}

w2

ς = ∞ X n=1 2n − 3!! 2n−1_n! [ ∞ X m=1 2m−1( r R + r) m_]2n−1_.

„p

_:

*

_(2.61)

û˝6ªø

r R − r = r R + r(1 − 2r R + r) −1 (2.66) = r R + r ∞ X m=0 ( 2r R + r) m _(2.67) = ∞ X m=1 2m−1( r R + r) m_{. (2.68)}

Žâ

(2.64)

¸

_(2.68)

û˝6ª„|

(2.66).

úı

3b!‹

ø

_{^:sR.,ä5q„p}

…ðÊ„p

s2 _{≤ 4R}2_{+ 4Rr + 3r}2_,

_{„pFàƒíxX}

_,

₃

_b¡5SQj

_,

_i

Y\

,

-î•

,

Ù

−p

(2007)

5dı

,

vÌHà-

:

âk

4r2s2(4R2+ 4Rr + 3r2− s2_{) = a}2_b2_c2_{+ 4abc(s − a)(s − b)(s − c)} +12(s − a)2(s − b)2(s − c)2 −4s3_{(s − a)(s − b)(s − c)} = (X + Y )2(Y + Z)2(X + Z)2 +4(X + Y )(Y + Z)(X + Z)XY Z + 12X2Y2Z2 −4(X + Y + Z)3_{XY Z,}

]

4r2s2(4R2+ 4Rr + 3r2− s2_{) = (U + 2V )}2_{+ 4V (U + 2V ) + 12V}2 −4V (X3_{+ Y}3_{+ Z}3_{+ 3U + 6V )} = U2− 4V (X3+ Y3+ Z3) − 4U V = X4(Y + Z)2+ Y4(X + Z)2+ Z4(X + Y )2 +2X2Y2(Y + Z)(X + Z) + 2Y2Z2(X + Z)(X + Y ) +2X2Z2(Y + Z)(X + Y ) − 4XY Z(X3+ Y3+ Z3) −4XY Z(X2_{Y + XY}2_{+ X}2_{Z + XZ}2_{+ Y}2_{Z + Y Z}2_),

w2

_,

I

_{U = X}2_{Y + XY}2_{+ X}2_{Z + XZ}2_{+ Y}2_{Z + Y Z}2_{, V = XY Z}

;WXÌb×kkSÌb

,

û

˝6

X4(Y + Z)2+ Y4(X + Z)2+ Z4(X + Y )2+ 2X2Y2(Y + Z)(X + Z) +2Y2Z2(X + Z)(X + Y ) + 2X2Z2(Y + Z)(X + Y ) − 4XY Z(X3+ Y3+ Z3) −4XY Z(X2Y + XY2+ X2Z + XZ2+ Y2Z + Y Z2) ≥ 4X4_{Y Z + 4XY}4_{Z + 4XY Z}4_{+ 2X}2_Y2_{(Y + Z)(X + Z)} +2Y2Z2(X + Z)(X + Y ) + 2X2Z2(Y + Z)(X + Y ) − 4XY Z(X3+ Y3+ Z3) −4XY Z(X2Y + XY2+ X2Z + XZ2+ Y2Z + Y Z2),

¥U)

4r2s2(4R2+ 4Rr + 3r2− s2) = 2[(X2Y2Z2− X2Y3Z − X3Y2Z + X3Y3) +(X2Y2Z2− XY3_Z2 _{− XY}2_Z3_{+ Y}3_Z3₎ +(X2Y2Z2− X2_{Y Z}3 _{− X}3_{Y Z}2_{+ X}3_Z3_)] = 2[X2Y2(Y − Z)(X − Z) + Y2Z2(X − Z)(X − Y ) −X2_Z2_{(Y − Z)(X − Y )] ≥ 0.}

Ĥ)„

s2 ≤ 4R2+ 4Rr + 3r2.

ù

_{^:sR.-ä5q„p}

…

ðÊ„p

16Rr − 5r2 _{≤ s}2_,

_„pqñà-

_:

_I

_{X = s − a, Y = s − b, Z =} s − c,

.Üø

O4

,

cq

X ≥ Y ≥ Z

FJ

X(X − Y )2+ Z(Y − Z)2+ (X − Y )(Y − Z)(X − Y + Z) ≥ 0.

ÇøjÞ

,

ÄÑ

s(s2− 16Rr + 5r2_{) = s(2s}2_{− 8Rr − 2r}2_{) − (s}2_{+ 4Rr + r}2_{) + 8r}2_{− 4Rr}

= s(a2+ b2+ c2) − (ab + bc + ca) + 8r2− 4Rr

= s(a2+ b2+ c2) − s(ab + bc + ca)

+(b + c − a)(a + b − c)(a + c − b) − abc

= sb2− 2abs + sa2− ab2+ 2a2b − a3+ sc2− 2bcs

+sb2− b2_{c + 2bc}2_{− c}3_{+ abs + 2ab}2_{− a}2_{b + bcs}

−sb2_{+ 2b}2_{c − bc}2_{+ 2ac}2_{− 3abc − b}3_{− acs}

= (s − a)(b − a)2+ (s − c)(c − b)2

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+(b2c − b3− abc + ab2_{) − (bc}2_{− b}2_{c − ac}2_{+ abc),}

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a b c s s − a s − b s − c R r s2_{− 16Rr + 5r}2 48 48 49 72.5 24.5 24.5 23.5 27.909338 13.948613 0.324137 49 49 50 74 25 25 24 28.486604 14.23737 0.324324 50 50 51 75.5 25.5 25.5 24.5 29.063873 14.526124 0.324503 51 51 52 77 26 26 25 29.641145 14.814875 0.324675 52 52 53 78.5 26.5 26.5 25.5 30.218421 15.103623 0.324841 53 53 54 80 27 27 26 30.795699 15.392368 0.325 54 54 55 81.5 27.5 27.5 26.5 31.372981 15.681111 0.325153 55 55 56 83 28 28 27 31.950264 15.969851 0.325301 56 56 57 84.5 28.5 28.5 27.5 32.52755 16.258589 0.325444 57 57 58 86 29 29 28 33.104838 16.547325 0.325581 58 58 59 87.5 29.5 29.5 28.5 33.682129 16.836058 0.325714 59 59 60 89 30 30 29 34.259421 17.12479 0.325842 60 60 61 90.5 30.5 30.5 29.5 34.836716 17.41352 0.325966 61 61 62 92 31 31 30 35.414012 17.702247 0.326087 62 62 63 93.5 31.5 31.5 30.5 35.99131 17.990974 0.326203 63 63 64 95 32 32 31 36.56861 18.279698 0.326315 64 64 65 96.5 32.5 32.5 31.5 37.145911 18.568421 0.326424 65 65 66 98 33 33 32 37.723214 18.857143 0.326531

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a b c s s − a s − b s − c R r s2_{− 16Rr + 5r}2 66 66 67 99.5 33.5 33.5 32.5 38.300519 19.145863 0.326633 67 67 68 101 34 34 33 38.877824 19.434582 0.326732 68 68 69 102.5 34.5 34.5 33.5 39.4551314 19.723299 0.326829 69 69 70 104 35 35 34 40.032439 20.012016 0.326923 70 70 71 105.5 35.5 35.5 34.5 40.609749 20.300731 0.327014 71 71 72 107 36 36 35 41.187059 20.589445 0.327028 72 72 73 108.5 36.5 36.5 35.5 41.764371 20.878158 0.327188 73 73 74 110 37 37 36 42.341685 21.166869 0.327272 74 74 75 111.5 37.5 37.5 36.5 42.918998 21.45558 0.327354 75 75 76 113 38 38 37 43.496313 21.74429 0.327433 76 76 77 114.5 38.5 38.5 37.5 44.073629 22.032999 0.327511 77 77 78 116 39 39 38 44.650946 22.321707 0.327586 78 78 79 117.5 39.5 39.5 38.5 45.228263 22.610415 0.327659 79 79 80 119 40 40 39 45.805582 22.899121 0.327731 80 80 81 120.5 40.5 40.5 39.5 46.382901 23.187827 0.327801 81 81 82 122 41 41 40 46.960221 23.476532 0.327869 82 82 83 123.5 41.5 41.5 40.5 47.537542 23.765236 0.327935 83 83 84 125 42 42 41 48.114863 24.053939 0.328

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a b c s s − a s − b s − c R r s2_{− 16Rr + 5r}2 22 22 21 32.5 10.5 10.5 11.5 12.517706 6.245921 0.353846 23 23 22 34 11 11 12 13.094701 6.534973 0.352941 24 24 23 35.5 11.5 11.5 12.5 13.671724 6.823994 0.352112 25 25 24 37 12 12 13 14.248773 7.112987 0.351351 26 26 25 38.5 12.5 12.5 13.5 14.825845 7.401956 0.350649 27 27 26 40 13 13 14 15.402936 7.690903 0.35 28 28 27 41.5 13.5 13.5 14.5 15.980045 7.979831 0.349397 29 29 28 43 14 14 15 16.557171 8.268741 0.348837 30 30 29 44.5 14.5 14.5 15.5 17.134311 8.557636 0.348314 31 31 30 46 15 15 16 17.711465 8.846517 0.347826 32 32 31 47.5 15.5 15.5 16.5 18.28863 9.135385 0.347368 33 33 32 49 16 16 17 18.865806 9.424241 0.346938 34 34 33 50.5 16.5 16.5 17.5 19.442993 9.713087 0.346534 35 35 34 52 17 17 18 20.020189 10.001923 0.346153 36 36 35 53.5 17.5 17.5 18.5 20.597392 10.290749 0.345794 37 37 36 55 18 18 19 21.174604 10.579568 0.345454 38 38 37 56.5 18.5 18.5 19.5 21.751823 10.868379 0.345132 39 39 38 58 19 19 20 22.329049 11.157184 0.344827 40 40 39 59.5 19.5 19.5 20.5 22.90628 11.445982 0.344537 41 41 40 61 20 20 21 23.483517 11.734773 0.3442622

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a b c s s − a s − b s − c R r s2_{− 16Rr + 5r}2 42 42 41 62.5 20.5 20.5 21.5 24.06076 12.02356 0.344 43 43 42 64 21 21 22 24.639001 12.312341 0.34375 44 44 43 65.5 21.5 21.5 22.5 25.215259 12.601117 0.343511 45 45 44 67 22 22 23 25.792516 12.889885 0.343283 46 46 45 68.5 22.5 22.5 23.5 26.369776 13.178657 0.343065 47 47 46 70 23 23 24 26.947041 13.467421 0.342857 48 48 47 71.5 23.5 23.5 24.5 27.524308 13.756181 0.342657 49 49 48 73 24 24 25 28.101579 14.044937 0.342465 50 50 49 74.5 24.5 24.5 25.5 28.678853 14.33369 0.342281 51 51 50 76 25 25 26 29.256131 14.622441 0.342105 52 52 51 77.5 25.5 25.5 26.5 29.833411 14.911188 0.341935 53 53 52 79 26 26 27 30.410692 15.199933 0.341772 54 54 53 80.5 26.5 26.5 27.5 30.987977 15.488675 0.341614 55 55 54 82 27 27 28 31.565265 15.777415 0.341463 56 56 55 83.5 27.5 27.5 28.5 32.142554 16.066152 0.341317 57 57 56 85 28 28 29 32.719846 16.354887 0.341176 58 58 57 86.5 28.5 28.5 29.5 33.297139 16.643621 0.34104 59 59 58 88 29 29 30 33.874435 16.932352 0.340909 60 60 59 89.5 29.5 29.5 30.5 34.451732 17.221081 0.340782 61 61 60 91 30 30 31 35.029031 17.509801 0.340659

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