A triangular inequality including squares of sides

Question

Show that for any triangle ABC, the following inequality is true

$$a^2 + b^2 + c^2 > \sqrt{3} \max\{|a^2-b^2|,|b^2-c^2|,|c^2-a^2|\}$$ where $a,b,c$ are the sides of the triangle

Answer 1

Let $c^2-a^2$ be the maximal term. By triangle inequality, we have $b^2>a^2-2ac+c^2$. Substituting in the equation of the OP, we get

$$a^2 + a^2 -2ac+ c^2 + c^2 > \sqrt{3} (c^2-a^2)$$

which leads to

$$ \frac{1}{2}[(2+\sqrt{3})a^2+(2-\sqrt{3})c^2]>ac $$

This last inequality is always true because it corresponds to an AM-GM inequality.