Cho $a,b,c>0$ thỏa mãn $a^2+b^2+c^2=3$. Chứng minh rằng:
$\frac{2a^2}{a+b^2}+\frac{2b^2}{b+c^2}+\frac{2c^2}{c+a^2}\geq a+b+c$
Cho $a,b,c>0$ thỏa mãn $a^2+b^2+c^2=3$. Chứng minh rằng:
$\frac{2a^2}{a+b^2}+\frac{2b^2}{b+c^2}+\frac{2c^2}{c+a^2}\geq a+b+c$
Cho $a,b,c>0$ thỏa mãn $a^2+b^2+c^2=3$. Chứng minh rằng:
$\frac{2a^2}{a+b^2}+\frac{2b^2}{b+c^2}+\frac{2c^2}{c+a^2}\geq a+b+c$
Theo bất đẳng thức $\text{AM-GM}$ ta có:
\[\sum \frac{a^2}{a+b^2} = \sum \left (a - \frac{ab^2}{a+b^2} \right )\ge \sum \left ( a-\frac{\sqrt{a}b}{2} \right ) \ge \frac{a+b+c}{2}\]
Edited by Mr Cooper, 20-05-2017 - 10:53.
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