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.

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## Comments

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TopNewestApply one of the variation of Heron's formula:

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

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 \(a=b=c\).

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