# Triangle Inequality

Suppose there is a triangle with sides $$a,b$$ and $$c$$. Prove that

$a^2 + b^2 + c^2 \geq \sqrt{48} \times A$

where $$A$$ denote the area of the triangle.

Note by Harmanjot Singh
2 years, 5 months ago

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## Comments

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$A = \frac14 \sqrt{ (a^2 +b^2+c^2)^2 - 2(a^4+b^4+c^4) }$

So we have to prove that $$a^2 + b^2+ c^2 \geq \dfrac{\sqrt{48}}4 \sqrt{(a^2+ b^2+c^2)^2 - 2(a^4 + b^4 + c^4) }$$.

Equivalently, (squaring both sides), we want to prove that $$16(a^2 + b^2+c^2) \geq 48( (a^2+b^2+c^2)^2 - 2(a^4 + b^4 + c^4) )$$ or $$(a^2 + b^2 + c^2)^2 \leq 3(a^4 + b^4 + c^4 )$$ which is obviously true when we apply the power mean inequality on the numbers $$a^2 , b^2, c^2$$.

$\text{QM}(a^2, b^2,c^2) \geq \text{AM}(a^2,b^2,c^2) \Rightarrow \sqrt[4]{\frac{a^4 + b^4+c^4}{3} } \geq \sqrt{ \frac{a^2+b^2 +c^2}3 }$

Equality holds when the triangle is equilateral, or $$a=b=c$$.

- 2 years, 5 months ago

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