cả nhà giúp với!!!!!!!!!!!!

Question

Cho tam giác ABC, CMR :

$\frac{sinA}{tan\frac{B}{2}} + \frac{sinB}{tan\frac{C}{2}} + \frac{sinC}{tan\frac{A}{2}} \geqslant \frac{9}{2}$

♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ · Answer 1 · 8/3/2014 12:19:54 PM

$bpt<=>(a+b+c)^4\ge9(ab+bc+ca)^2$

$<=>(a+b+c)^2\ge3(ab+bc+ca)$

$<=>a^2+b^2+c^2-ab-bc-ca\ge 0$

$<=>2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0$

$<=>(a-b)^2+(b-c)^2+(c-a)^2\ge0(ld)$

$=>dpcm$

$a,b,c$ luôn dơơng nên khi lấy căn không cần để ý tới dấu đâu

Trả lời 03-08-14 12:19 PM

♥♥♥ Panda Sơkiu Panda Mập ♥♥♥
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tieutulitipro · Answer 2 · 8/2/2014 10:58:14 PM

$2\frac{sinA}{tan\frac{B}{2}} = \frac{2sinA}{sinB}(1 + cosB) = 2\frac{\frac{a}{2R}}{\frac{b}{2R}}(1+\frac{a^2 + c^2 - b^2}{2ac})$

$= 2\frac{a}{b}[\frac{(a+c)^2-b^2}{2ac}]= \frac{(a+b+c)(a+c-b)}{bc}$ (1)

tương tự:

$2\frac{sinB}{tan\frac{C}{2}}= \frac{(a+b+c)(a+b-c)}{ac}$ (2)

$2\frac{sinC}{tan\frac{A}{2}}= \frac{(a+b+c)(b+c-a)}{ab}$ (3)

cộng lại dùng BĐT cộng mẫu ta được:

$(1)+(2)+(3) \geqslant \frac{(a+b+c)^4}{ab+bc+ca}\geqslant9(ab+bc+ca)$ , tới đây em potax

Trả lời 02-08-14 10:58 PM

tieutulitipro
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ai giup minh giai với – tieutulitipro 03-08-14 07:54 AM

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