If $\frac {1}{p}+\frac {1}{q}=1$ and $a,b \ge 0$ , then prove $ab\le \frac{a^p}{p}+\frac{b^q}{q}$ . I can't find a simple and short way to prove this. Any hint would work. Thanks in advance!
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3This is called Young's inequality. It is equivalent to showing $a^{1/p}b^{1/q} \le \frac1{p} a + \frac1{q}b$, which follows from the convexity of $\ln$ (supposing that $a,b > 0$; the other case is trivial) – Jun 26 '16 at 05:20
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1this is the inequality of Young – Dr. Sonnhard Graubner Jun 26 '16 at 05:20
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@ahemd thank you very much for your hint. – Vineet Mangal Jun 26 '16 at 05:30
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you can prove this using the AM-GM inequality as well – clark Jun 26 '16 at 05:30
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@clark I was trying but i wan not able to prove. Can you explain how ? – Vineet Mangal Jun 26 '16 at 05:31
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@VineetMangal You prove this firstly for $p,q$ rational numbers and then you say it is true for every number by density – clark Jun 26 '16 at 05:33
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I think the proof here is short and elegant! – awllower Jun 26 '16 at 05:33
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@awllower that was a wonderful proof. But i also got one good proof on wikipedia. Thank you very much for your help. – Vineet Mangal Jun 26 '16 at 05:38
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@VineetMangal that proof does exactly what I suggested – clark Jun 26 '16 at 05:41
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@clark thank you very much for your help. – Vineet Mangal Jun 26 '16 at 05:43
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Some other related posts: http://math.stackexchange.com/questions/149901/geometric-interpretation-of-youngs-inequality and http://math.stackexchange.com/questions/259102/youngs-inequality-without-using-convexity – Martin Sleziak Jun 26 '16 at 07:10
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The general theorem of means (perhaps more popularly known as Young's inequality as the comment to this answer suggests) implies that if $x, y, r, s$ are positive with $r + s = 1$ then $$x^{r}y^{s} \leq rx + sy\tag{1}$$ Putting $x = a^{p}, y = b^{q}, r = 1/p, s = 1/q$ we get $$ab \leq \frac{a^{p}}{p} + \frac{b^{q}}{q}$$
Paramanand Singh
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-1 It is called Young's inequality, not general theorem of means; the latter term does not show up in any google search. Please do not use non-standard terminology when referring to theorems (If I am wrong tell me and I'll happily revert my downvote) – Ant Jun 26 '16 at 07:29
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@Ant The inequality he refers to is generally called the weighted AM-GM inequality. I guess the other name is a translation of the name in the OP's native language. (It would probably be a good idea to change this in the post) – wythagoras Jun 26 '16 at 07:51
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1@Ant: Thanks for your comment. I strongly dislike the fashion of giving downvote without comment and you are an exception to this. It means a great deal here. The terminology is not standard and I have got this from Hardy's "A Course of Pure Mathematics". I will add your terminology also in the answer. – Paramanand Singh Jun 26 '16 at 08:04
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@Ant: Its so bad everyone here is not like you. I have now received a downvote without any explanation (and yes I know you had reversed yours). That's why I said your comment means a great deal. I wonder why people don't explain their downvote. – Paramanand Singh Jun 26 '16 at 11:42
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Would the downvoter care to comment and learn something from users like Ant? – Paramanand Singh Jun 26 '16 at 11:45