oVlad wrote: Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2\geqslant 3$. Prove that \[\frac{a^2}{a+b^2}+\frac{b^2}{b+c^2}+\frac{c^2}{c+a^2}\geqslant\frac{3}{2}.\] Let $x,y,z$ be positive real numbers satisfying $x+y+z=3$.Prove that: $$ \frac{x^2}{x+y^2}+\frac{y^2}{y+z^2}+\frac{z^2}{z+x^2} \ge \frac{3}{2}$$

oVlad wrote: Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2\geqslant 3$. Prove that \[\frac{a^2}{a+b^2}+\frac{b^2}{b+c^2}+\frac{c^2}{c+a^2}\geqslant\frac{3}{2}.\] Sines $$\sum_{cyc}\frac{a^2}{a+b^2}\geq\sum_{cyc}\frac{a^2}{a\sqrt{\frac{a^2+b^2+c^2}{3}}+b^2},$$we can assume $a^2+b^2+c^2=3,$ which by C-S gives: $$\sum_{cyc}\frac{a^2}{a+b^2}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^3+a^2b^2)}=\frac{9}{\sum\limits_{cyc}\left(a^3+\frac{3-a^4}{2}\right)}=\frac{3}{2}+\frac{9}{\sum\limits_{cyc}\left(a^3+\frac{3-a^4}{2}\right)}-\frac{3}{2}=$$$$=\frac{3}{2}+\frac{36-3\sum\limits_{cyc}(2a^3+3-a^4)}{2\sum\limits_{cyc}(2a^3+3-a^4)}=\frac{3}{2}+\frac{3\sum\limits_{cyc}(a^4-2a^3+a^2)}{2\sum\limits_{cyc}(2a^3+3-a^4)}=\frac{3}{2}+\frac{3\sum\limits_{cyc}a^2(a-1)^2}{2\sum\limits_{cyc}(2a^3+3-a^4)}\geq\frac{3}{2}.$$

How is this equal? \[\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^3+a^2b^2)}=\frac{9}{\sum\limits_{cyc}\left(a^3+\frac{3-a^4}{2}\right)}\]

youthdoo wrote: How is this equal? \[\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^3+a^2b^2)}=\frac{9}{\sum\limits_{cyc}\left(a^3+\frac{3-a^4}{2}\right)}\] 1) $a^2+b^2+c^2=3$, so $(a^2+b^2+c^2)^2=9$ 2) $2\sum a^2b^2 + \sum a^4 = (a^2+b^2+c^2)^2=9=3+3+3,$ so $2\sum a^2b^2 = \sum(3-a^4)$ or $\sum a^2b^2=\sum \frac{3-a^4}{2}$

I think it is simply $\ge$ not $=$...

youthdoo wrote: I think it is simply $\ge$ not $=$... No, there is strict equality everywhere (in #5)

But it is $a^2+b^2+c^2\ge3$, not $a^2+b^2+c^2=3$.

arqady wrote: we can assume $a^2+b^2+c^2=3$