17th Junior Turkish Mathematical Olympiad

Let $f(a,b,c)=\frac{a}{a^2+b^3+c^3}+\frac{b}{a^3+b^2+c^3}+\frac{c}{a^3+b^3+c^2}$ and $g(a,b,c)=a(a^2+b^3+c^3)+b(a^3+b^2+c^3)+c(a^3+b^3+c^2)$. Observe that the Cauchy-Schwarz Inequality gives $f(a,b,c)\cdot g(a,b,c)\ge (a+b+c{)}^2$.

Observe that $g(a,b,c)=(a+b+c)(a^3+b^3+c^3)$ since $a^3+b^3+c^3=a^4+b^4+c^4$ and therefore it suffices to show that $a+b+c\ge a^3+b^3+c^3$. Again by the Cauchy-Schwarz Inequality we have $(a+b+c)(a^3+b^3+c^3)\ge (a^2+b^2+c^2{)}^2$ and $(a^2+b^2+c^2)(a^4+b^4+c^4)\ge (a^3+b^3+c^3{)}^2$. As $a^3+b^3+c^3=a^4+b^4+c^4$, the result follows from the last two inequalities.