<s>[INST] How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$? How can the following inequation be proven?
$$a^2 + b^2 + c^2 \ge ab + bc + ca$$
 [/INST] This inequality can be solved by simple algebra
we have the equation
$$a^2+b^2+c^2 >= ab+ac+cb$$
multiply and divide 2 on both sides
$$\frac{2}{2}(a^2+b^2+c^2) >= \frac{2}{2}(ab+ac+cb)$$
putting the left side equation on the right side
$$\frac{2a^2+2b^2+2c^2 - 2(ab+ac+cb)}{2} >= 0$$
then factorize and multiply by 2 on both the sides
$$\frac{2a^2+2b^2+2c^2 - 2(ab+ac+cb)}{2} >= 0$$
$$\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2} >= 0$$
$$\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2} >= 0$$
$$2*\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2} >= 2*0$$
$$(a-b)^2+(b-c)^2+(c-a)^2 >= 0$$
So the above equation is zero when $a=b=c$
So the above inequality is proved
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