Inequality 1

[MM]

Prove for any positive numbers a,b,c
[tex]\frac{a^{3}}{b^{2}+bc+c^{2}}+\frac{b^{3}}{c^{2}+ca+a^{2}}+\frac{c^{3}}{a^{2}+ab+b^{2}}\ge \frac{ab+bc+ca}{a+b+c}[/tex]

[Rock'n'roller]

[tex]\frac{a^4}{ab^2 + abc + ac^2} + \frac{b^4}{bc^2 + abc + ba^2} + \frac{c^4}{ca^2 + abc + cb^2} \ge \frac{(a^2+b^2+c^2)^2}{a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 + 3abc} \ge \frac{(ab + bc + ca)^2}{a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 + 3abc}[/tex]

[tex]\frac{(ab + bc + ca)^2}{a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 + 3abc} \ge \frac{ab+bc+ca}{a+b+c} \Leftrightarrow \frac{ab+bc+ca}{a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 + 3abc} \ge \frac{1}{a+b+c} \Leftrightarrow 0 \ge 0[/tex].

I'm not sure

[MM]

Right!