\[\dfrac{a}{b+2c+3d} + \dfrac{b}{c+2d+3a} + \dfrac{c}{d+2a+3b} + \dfrac{d}{a+2b+3c} > \dfrac{2}{3}\]

If a,b,c,d are distinct positive reals, prove the above inequality.

\[\dfrac{a}{b+2c+3d} + \dfrac{b}{c+2d+3a} + \dfrac{c}{d+2a+3b} + \dfrac{d}{a+2b+3c} > \dfrac{2}{3}\]

If a,b,c,d are distinct positive reals, prove the above inequality.

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TopNewestHave u given the inmo mock test of fiitjee yesterday – Samarth Agarwal · 1 year ago

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– Raushan Sharma · 1 year ago

No, one of my friends gave that, and yeah, it's a question I gave from there only... Did you give that test?? How was it?Log in to reply

– Samarth Agarwal · 1 year ago

Yes I gave the test and it was really difficult....this was the only question I could completely solve....I would post the complete paper soon.Log in to reply

– Raushan Sharma · 1 year ago

Ya, I have the complete paper, but can you give the solution to this inequality, I mean how you solved, I was trying it with Tittu's Lemma, but couldn't completeLog in to reply

– Samarth Agarwal · 1 year ago

Use titu lemma and then cauchy schwarz on \(sqrt (a), sqrt (b), sqrt (c), sqrt (d) and 1,1,1,1\) it would give the direct resultLog in to reply

– Raushan Sharma · 1 year ago

Oh, yeah, I did it today, after I commented that. It was quite easy. Actually first I was not expanding \((a+b+c+d)^2\). I first applied Tittu's lemma and then AM-GMLog in to reply

Can you please post the complete solution including explanation for this question??– Saurabh Mallik · 10 months, 3 weeks agoLog in to reply

– Raushan Sharma · 10 months, 3 weeks ago

Yes, I can, but for that can you please tell me, how can we add an image in the comment??Log in to reply

Sorry. I don't know how can we upload images in comment. I think we can't upload images in comment but and upload it in solutions for given questions. Can you please write the whole solution?– Saurabh Mallik · 10 months, 3 weeks agoLog in to reply