$\frac{1}{4a^2-ab+4b^2}+\frac{1}{4b^2-bc+4c^2}+\frac{1}{4c^2-ca+4a^2}\geq \frac{9}{7(a^2+b^2+c^2)}$
Edited by no matter what, 29-12-2012 - 13:01.
Edited by duongtoi, 12-12-2012 - 15:07.
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Áp dụng bất đẳng thức Cauchy ta có,
$ab\le\frac{a^2+b^2}{2}$
Do vậy ta có,
$VT\ge \frac{2}{7(a^2+b^2)}+\frac{2}{7(c^2+b^2)}+\frac{2}{7(a^2+c^2)}\ge \frac{2}{7}3.\sqrt[3]{\frac{1}{(a^2+b^2)(c^2+b^2)(a^2+c^2)}}$
$\ge \frac{2}{7}3.\frac{1}{\frac{(a^2+b^2)+(b^2+c^2)+(c^2+a^2)}{3}}=\frac{9}{7(a^2+b^2+c^2)}$
Dấu $"="$ xảy ra khi và chỉ khi $a=b=c$
Edited by banhgaongonngon, 12-12-2012 - 15:15.
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