<s>[INST] 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.
 [/INST] Consider the polynomial $(x-a)(x-b)(x-c)$. Multiplied out, this is
$$
\begin{eqnarray}
(x-a)(x-b)(x-c)
&=&
x^3-(a+b+c)x^2+(ab+ac+bc)x-abc
\\
&=&
x^3-10x^2+(ab+ac+bc)x-abc\;.
\end{eqnarray}
$$
We also have
$$10^2=(a+b+c)^2=(a^2+b^2+c^2) + 2(ab+ac+bc)\;,$$
$$ab+ac+bc=\frac{100-(a^2+b^2+c^2)}{2}\;,$$
and thus
$$
(x-a)(x-b)(x-c)
=
x^3-10x^2+\frac{100-(a^2+b^2+c^2)}{2}x-abc\;.
$$
Now since $a$, $b$ and $c$ form a triangle with perimeter $10$, they must all be less than $5$. Thus the value of the polynomial for $x=5$ is positive, that is,
$$
5^3-10\cdot5^2+\frac{100-(a^2+b^2+c^2)}{2}\cdot5-abc>0\;,
$$
which upon rearrangement becomes your inequality.
</s>