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.

#Geometry #TriangleInequality

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Comments

Apply 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 $a=b=c$ .

Pi Han Goh - 5 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$