# University level Geometry question

Show that for any triangle ABC, the following inequality is true
a^2 + b^2 + c^2 > (3)^1/2 max [ | a^2 - b^2 |, | b^2 - c^2 |, | c^2 - a^2 | ]

Where a,b,c are the sides of the triangle in the usual notation.

(I tried using Law of cosine to solve it but I wasn't able to simplify the equations using that, I think inequality can be used to solve this question but my geometry is rusty )

Without loss of generality we may assume $a \geq b \geq c$. Then
$\max \{ | a^2 - b^2 |, | b^2 - c^2 |, | c^2 - a^2 | \}=a^2-c^2.$
It is enough to show that
$a^2+b^2+c^2 = \sqrt{3}(a^2-c^2) (*).$
By law of cosines we have
$$a^2+b^2+c^2 \geq a^2+(a^2+c^2-2ac \cos \theta)+c^2$$
$\geq a^2+(a^2+c^2-2ac )+c^2=2(a^2+c^2-ac)$
$\geq \sqrt{3}(a^2-c^2)+(2-\sqrt{3})a^2+(2+\sqrt{3})c^2-2ac$
$\geq \sqrt{3}(a^2-c^2)+(\sqrt{(2-\sqrt{3})}a-\sqrt{(2+\sqrt{3})}c)^2$
$\geq \sqrt{3}(a^2-c^2).$
Hence
$a^2+b^2+c^2 \geq \sqrt{3}(a^2-c^2).$

• Aaru
+1

Thank you so much for helping me out with the solution Paul:)

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