Prove 16(a²+b²+c²)−5(a+b+c)² ≥ 0 and find its smallest positive value

[anemone]

Let $a,\,b$ and $c$ be integers, and given $X=16(a^2+b^2+c^2)-5(a+b+c)^2$, prove that $X\ge 0$ and find the smallest positive number that $X$ can be.

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[anemone]

Congratulations to the following members for their correct solutions::)

1. kaliprasad
2. greg1313

Solution from kaliprasad:

Spoiler

$\begin{align*}X&= 16(a^2+b^2+c^2) - 5(a+b+c)^2\\&=16(a^2+b^2+c^2) - 5(a^2+b^2+c^2+2ab+2ac+2bc)\\&=11(a^2+b^2+c^2) - 10(ab+bc+ca)\\&=a^2+b^2+c^2 + 5(2a^2+2b^2+2c^2-2ab - 2bc - 2ca)\\&=(a^2+b^2+c^2 + 5((a-b)^2+(b-c)^2 + (c-a)^2)\end{align*}$

Clearly the lowest value is zero when $a = b= c = 0$

If $a,\,b,\,c$ are not same then the lowest value is 5 and if same and 1 or -1 then the value is 3.

So lowest positive value = 3 at $a=b= c = 1$ or $a=b=c = -1$.