@Pi Han Goh
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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
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Subham Subian
·
1 week, 4 days ago

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@Subham Subian
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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.
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Pi Han Goh
·
1 week, 3 days ago

@Pi Han Goh
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it was in a book(challenge and thrills of pre college mathematics) but i write the question write
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Subham Subian
·
1 week, 3 days 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.
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Prasun Biswas
·
1 week, 2 days ago

## Comments

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TopNewestBy rearrangement inequality, \(a^2 + b^2 + c^2 \geq ab + ac + bc\), then apply Weitzenböck's inequality, the result follows. – Pi Han Goh · 1 week, 4 days ago

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– Subham Subian · 1 week, 4 days ago

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^2Log in to reply

– Pi Han Goh · 1 week, 3 days ago

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.Log in to reply

\(\large\ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }\ge 4\sqrt { 3 } X\) – Priyanshu Mishra · 1 week, 3 days ago

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– Priyanshu Mishra · 1 week, 3 days ago

It is evident from Paul-Erdos InequalityLog in to reply

– Subham Subian · 1 week, 3 days ago

it was in a book(challenge and thrills of pre college mathematics) but i write the question writeLog in to reply

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 week, 2 days ago

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– Calvin Lin Staff · 6 days, 19 hours ago

Thanks! I've reversed the sign.Log in to reply