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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.

- 1 year, 3 months 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^2

- 1 year, 3 months 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.

- 1 year, 3 months ago

The correct question is to prove;

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

- 1 year, 3 months ago

It is evident from Paul-Erdos Inequality

- 1 year, 3 months ago

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

- 1 year, 3 months ago

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.

- 1 year, 3 months ago

Thanks! I've reversed the sign.

Staff - 1 year, 3 months ago