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Inequalities

If \(a,b\) and \(c\) are sides of a triangle \(ABC\) with area \(X\), prove that \(ab + bc + ca \geq 4 \sqrt3 X\).

Note by Subham Subian
1 year, 3 months ago

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By rearrangement inequality, \(a^2 + b^2 + c^2 \geq ab + ac + bc\), then apply Weitzenböck's inequality, the result follows.

Pi Han Goh - 1 year, 3 months ago

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ok but i have a doubt we know that a^2+b^2+c^2>=ab+bc+ca and a^2+b^2+c^2>=4(square root of 3)(X) but how u got the relation between these two since both are less than a^2+b^2+c^2

Subham Subian - 1 year, 3 months ago

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Are you sure you got your question right? After reading your question for the second time, I'm pretty sure your inequality sign should be reversed.

Pi Han Goh - 1 year, 3 months ago

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@Pi Han Goh The correct question is to prove;

\(\large\ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }\ge 4\sqrt { 3 } X\)

Priyanshu Mishra - 1 year, 3 months ago

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@Priyanshu Mishra It is evident from Paul-Erdos Inequality

Priyanshu Mishra - 1 year, 3 months ago

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@Pi Han Goh it was in a book(challenge and thrills of pre college mathematics) but i write the question write

Subham Subian - 1 year, 3 months ago

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Counterexample: Consider the 3-4-5 triangle for which we have \((a,b,c)=(3,4,5)\) and hence \(ab+bc+ca=12+20+15=47\) whereas the area \(X=\frac{3\times 4}2=6\) from which we have \(4\sqrt 3X=24\sqrt 3\approx 41.569\), so we have \(ab+bc+ca\gt 4\sqrt 3X\)


However, when the inequality sign is reversed, the claim is correct and the proof for that is given by Pi Han Goh in his comment.

Prasun Biswas - 1 year, 3 months ago

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Thanks! I've reversed the sign.

Calvin Lin Staff - 1 year, 3 months ago

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