Cho $a,b,c> 0$ . chứng minh:
$\frac{1}{1+2ab}+\frac{1}{1+2bc}+\frac{1}{1+2ca}+\frac{4(a+b+c)}{9}\geqslant \frac{7}{9}$
Ta có : $S=\frac{1}{1+2ab}+\frac{1}{1+2bc}+\frac{1}{1+2ac}$
=$3-\left ( \frac{2ab}{1+2ab}+\frac{2bc}{1+2bc} +\frac{2ac}{1+2ac}\right )$
$\geq 3-\left ( \frac{2ab}{3\sqrt[3]{a^{2}b^{2}}}+\frac{2bc}{3\sqrt[3]{b^{2}c^{2}}}+\frac{2ac}{3\sqrt[3]{a^{2}b^{2}}}\right )$
$=3-\frac{2}{3}\left ( \sqrt[3]{ab}+\sqrt[3]{bc} +\sqrt[3]{ac}\right )$
$\geq 3-\frac{2}{3}.\frac{2\left ( a+b+c \right )+3}{3}$
=$3-\frac{4\left ( a+b+c \right )}{9}-\frac{2}{3}$
Mà :$P=S+\frac{4\left ( a+b+c \right )}{9}$
$\Rightarrow P\geq 3-\frac{2}{3}=\frac{7}{3}\Rightarrow$ đpcm