Question

Let $a,b,c>0,a+b+c=15$.
The least value of $E=\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}$ is___

Answer 1

Observe that
$\frac{a^3-b^3}{a^2+ab+b^2}+\frac{b^3-c^3}{b^2+bc+c^2}+\frac{c^3-a^3}{c^2+ca+a^2}=(a-b)+(b-c)+(c-a)=0Thus,wecanwriteE=\frac{1}{2}(\frac{a^3+b^3}{a^2+ab+b^2}+\frac{b^3+c^3}{b^2+bc+c^2}+\frac{c^3+a^3}{c^2+ca+a^2})But\frac{a^3+b^3}{a^2+ab+b^2}\ge \frac{a+b}{3}Since{a}^2-2ab+b^2=(a-b{)}^2\ge 0\therefore E\ge \frac{1}{3}(a+b+c)=5$