[Toán 10]Chứng minh Bất đẳng thức

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ngomaithuy93

Cho a, b, c > 0 và a + b + c =1.

CMR: [TEX]\frac{a^2+b}{b+c}+\frac{b^2+c}{c+a}+\frac{c^2+a}{a+b} \geq 2[/TEX]
[TEX]* \frac{a^2}{b+c}+\frac{b+c}{4}+\frac{a}{2} \geq \frac{3a}{2}[/TEX]
[TEX]\Rightarrow \frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{1}{2}[/TEX]
[TEX] * \frac{b}{b+c}+\frac{3(b+c)}{4}+\frac{9b^2}{2} \geq \frac{9}{2}b[/TEX]
[TEX] \Rightarrow \frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b} \geq \frac{3}{2}[/TEX]
\Rightarrow dpcm
[TEX] "=" \Leftrightarrow a=b=c=\frac{1}{3}[/TEX]
 
D

dandoh221

Cho a, b, c > 0 và a + b + c =1.

CMR: [TEX]\frac{a^2+b}{b+c}+\frac{b^2+c}{c+a}+\frac{c^2+a}{a+b} \geq 2[/TEX]
Giả sử [TEX]a = max{a,b,c}[/tex]
[TEX]\sum \frac{a^2}{b+c} \ge \sum \frac{ac}{b+c}[/TEX]
[TEX]\Leftrightarrow \frac{a+b+c}{(b+c)(c+a)}(a-b)(a-c) + \frac{a+b+c}{(a+b)(a+c)}.(b-c)^2 \ge 0[/TEX] đúng
[TEX]\Rightarrow \frac{a^2+b}{b+c}+\frac{b^2+c}{c+a}+\frac{c^2+a}{a+b}\ge \sum \frac{ac+b}{b+c} = \sum \frac{ac+b(a+b+c)}{b+c} = 2[/TEX] :)
 
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