Chém luôn bài này:Bài 470: Cho a,b,c>0. Chứng minh:
$\frac{1}{6}(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac})^{2}\geq \sum \frac{1}{a^{4}+b^{2}c^{2}}$
Áp dụng BĐT Cauchy,ta có:
$\sum \frac{1}{a^4+b^2c^2}\leq \sum \frac{1}{2a^2bc}=\frac{1}{2abc}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$
Áp dụng BĐT $(a+b+c)^2\geq 3(ab+bc+ca)$,ta có:
$\frac{3}{abc}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\leq (\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})^2$
$=>\sum \frac{1}{a^4+b^2c^2}\leq \frac{1}{2abc}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\leq \frac{1}{6}(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})^2(Q.E.D)$