Ta c/m: $\Sigma \frac{a^2+b^2}{a+b}\geq \sqrt{3(\Sigma a^2)}$
Cách 1:
$\Sigma (\frac{a^2+b^2}{a+b}-\frac{a+b}{2})\geq \sqrt{3(\Sigma a^2)}-(\Sigma a)$
$\Leftrightarrow \Sigma \frac{(a-b)^2}{2(a+b)}\geq \frac{\Sigma (a-b)^2}{\sqrt{3(\Sigma a^2)}+(\Sigma a)}$
Mà:
$(a+b+c)^2\leq 3(a^2+b^2+c^2)$
nên:
$\sqrt{3(a^2+b^2+c^2)}+a+b+c\geq 2(a+b+c)$
Ta cần c/m:
$\Sigma \frac{(a-b)^2}{2(a+b)}\geq \frac{\Sigma (a-b)^2}{2(a+b+c)}$
$\Leftrightarrow \Sigma \frac{c(a-b)^2}{a+b}\geq 0$ lđ!
$\Rightarrow ..............$
Đẳng thức khi $a=b=c./$