$\bigstar$ VD 11:
$\sum \sqrt{\frac{a^2}{a^2+7ab+b^2}}\geq 1$
$\blacksquare \blacksquare \blacksquare$
$dpcm\Leftrightarrow \sum \frac{1}{\sqrt{1+7\frac{b}{a}+\frac{b^{2}}{a^{2}}}}\geq 1$
Đặt : $\sqrt{\frac{b}{a}}=x;\sqrt{\frac{c}{b}}=y;\sqrt{\frac{a}{c}}=z\Rightarrow xyz=1$
BĐT trở thành $\sum \frac{1}{\sqrt{x^{4}+7x^{2}+1}}\geq 1$
Ta chứng minh $\frac{1}{\sqrt{x^{4}+7x^{2}+1}}\geq \frac{1}{x^{2}+x+1}$ (1)
Thật vậy :
$(1)\Leftrightarrow (x^{2}+x+1)^{2}-(x^{4}+7x^{2}+1)\geq 0\Leftrightarrow 2x(x-1)^{2}\geq 0$ (luôn đúng)
$\Rightarrow \sum \frac{1}{\sqrt{x^{4}+7x^{2}+1}}\geq \sum \frac{1}{x^{2}+x+1}\geq 1$