<s>[INST] A triangular inequality including squares of sides 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
 [/INST] 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. 
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