I need your help to solve this Math Olympiad problem: Let $x$, $y$ and $z$ be positive real numbers such that $x^2 + y^2 + z^2 = x + y + z$.
Prove that : $$ \frac{x+1}{\sqrt{x^5+x+1}}+\frac{y+1}{\sqrt{y^5+y+1}}+\frac{z+1}{\sqrt{z^5+z+1}}\ge 3 $$ In order to solve that problem: I tried to prove that each fraction in the inequality is > or equal to 1 in the given case. So I factored the denominator of the expression: $$ \frac{x+1}{\sqrt{x^5+x+1}} $$ To make it equal to: $$ \sqrt{(x^2+x+1)(x^3-x^2+1)}\ $$ Then I used this relation: $$ \sqrt{ab} \le \frac{a + b}{2} $$ To conclude that: $$ \frac{x+1}{\sqrt{(x^2+x+1)(x^3-x^2+1)}} \ge \frac{2x + 2}{x^3+x+2} $$ The problem I faced is that I did not know how to link the given information that was presented to the result that I reached, and I do not really know if what I did was useful. In the end, I think I tried enough, and I need the opinions of those who are more knowledgeable than me. I would be grateful if someone could help me find a solution, thanks in advance.
You're almost there. Note $(x+1) \mid (x^3+x+2)$, so we have $$\frac{2x+2}{x^3+x+2} = \frac2{x^2-x+2}$$
Now looking at the denominator's $x^2-x$ term should remind you of the constraint; and CS inequality gives: $$2\sum_{cyc}\frac1{x^2-x+2} \geqslant 2\cdot\frac{9}{6+\sum_{cyc}(x^2-x)}=3$$