# 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.

-
what do u mean by a circumference of a triangle. –  M.Subramani May 15 '11 at 11:33
Presumably by "circumference" what's meant is what I'd call "perimeter". –  Gerry Myerson May 15 '11 at 12:11

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.

-
Bravo!${}{}{}{}$ –  Gerry Myerson May 15 '11 at 12:19