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