By C-S we have $\frac{a^2}{a^2+ab}+\frac{b^2}{b^2+bc}+\frac{c^2}{c^2+ca}\ge \frac{(a+b+c)^2}{a^2+b^2+c^2+ab+bc+ca}$,so it's enough to prove that $\frac{(a+b+c)^2}{a^2+b^2+c^2+ab+bc+ca}\ge \frac{a+b+c}{2}$,which is equivalent to $2(a+b+c)\ge a^2+b^2+c^2+ab+bc+ca=a+b+c+ab+bc+ca$.Thus it's enough to prove that $a+b+c=a^2+b^2+c^2\ge ab+bc+ca$,which is obvious.

Can you explain why is (a^2)/(a^2+bc)+(b^2)/(b^2+bc)+(c^2)/(c^2+ca)>= ((a+b+c)^2)/(a^2+b^2+c^2+ab+bc+ac). I can't understand first inequality . Please explain me.

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Tsima wrote: Can you explain why is (a^2)/(a^2+bc)+(b^2)/(b^2+bc)+(c^2)/(c^2+ca)>= ((a+b+c)^2)/(a^2+b^2+c^2+ab+bc+ac). I can't understand first inequality . Please explain me. It is called Titu's Lemma or C-S in Engel form.

Thanks!!!

Let $a$ , $b$ , $c$ be positive real numbers such that $a+b+c=a^2+b^2+c^2$ . Prove that $$\sum \frac{a^2}{b+c} \geq \sum \frac{a}{b+c}$$here

I found that problem really easily ( almost in 10 minutes )..... I dont think its an A3 problem... What do you think?

that is can solve by Chebyshev