Let$ a,b,c>0 $with $a^3+b^3+c^3=a^4+b^4+c^4$,Prove that
\[\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+c^3+a^3}+\frac{c}{c^2+a^3+b^3}\geq 1 \]
(2012 Turkey)
Proof:
by Cauchy-Schwarz,we have
\[\left(\sum{\frac{a}{a^2+b^3+c^3}}\right)\left[\sum{a(a^2+b^3+c^3)}\right]\geq (a+b+c)^2\]
Therefore,it's suffice to prove
\[(a+b+c)^2\geq \sum{a(a^2+b^3+c^3)}\]
Or
\[(a+b+c)^2\geq (a+b+c)(a^3+b^3+c^3)\]
\[\Leftrightarrow  (a+b+c)(a^4+b^4+c^4)^2\geq (a^3+b^3+c^3)^3\]

Done!