Proof of inequality I have problems with proving inequality : 

$${a^{2}}+b^2+c^2+\frac{2}{5}abc<50$$

where $a,b,c$ are the lengths of triangle's sides, and the circumference of the triangle is $10$.
Thanks.

We need to prove that
$$\frac{(a+b+c)(a^2+b^2+c^2)}{10}+\frac{2}{5}abc<\frac{(a+b+c)^3}{20}$$ or
$$(a+b+c)^3>8abc+2(a+b+c)(a^2+b^2+c^2)$$ or
$$\sum_{cyc}(a^3+3a^2b+3a^2c+2abc)>8abc+2\sum_{cyc}(a^3+a^2b+a^2c)$$ or
$$\sum_{cyc}\left(-a^3+a^2b+a^2c-\frac{2}{3}abc\right)>0$$ or
$$(a+b-c)(a+c-b)(b+c-a)>0$$
Done!
