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 )

Geometry
Aaru Aaru
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1 Answer

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
\[a^2+b^2+c^2 \geq \sqrt{3}(a^2-c^2).\]

Paul F Paul F
  • Aaru Aaru
    +1

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

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