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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
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=23×4=6 from which we have 43X=243≈41.569, so we have ab+bc+ca>43X
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.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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By rearrangement inequality, a2+b2+c2≥ab+ac+bc, then apply Weitzenböck's inequality, the result follows.
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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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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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a2+b2+c2≥43X
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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=23×4=6 from which we have 43X=243≈41.569, so we have ab+bc+ca>43X
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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Thanks! I've reversed the sign.