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OK guys I have this problem:

For $x,y,p,q>0$ and $ \frac {1} {p} + \frac {1}{q}=1 $ prove that $ xy \leq\frac{x^p}{p} + \frac{y^q}{q}$

It says I should use Jensen's inequality, but I can't figure out how to apply it in this case. Any ideas about the solution?

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  • $\begingroup$ $\log$ is a concave function, so use that on RHS. $\endgroup$ – Cortizol Feb 14 '14 at 22:07
  • $\begingroup$ İ lookedİ am so so so sorry.for this question.İ can not delete this question.I do not have authority $\endgroup$ – user442995 Jun 3 '17 at 21:05
  • $\begingroup$ Also this might be worth looking at: Geometric interpretation of Young's inequality $\endgroup$ – Martin Sleziak Jun 5 '17 at 4:53
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The exponential funtion $t\mapsto \exp(t)$ is convex, so $$xy=\exp(\log(xy))=\exp(\log(x)+\log(y))=\exp((1/p)\log(x^p)+(1/q)\log(y^q))$$ $$\leq (1/p)\exp(\log(x^p))+(1/q)\exp(\log(y^q))=\frac{x^p}{p}+\frac{y^q}{q}.$$

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  • $\begingroup$ Upon closer inspection, I see that my answer is similar. If you wish I will delete my answer, but I think the presentations are different enough that the two might appeal to different people. $\endgroup$ – robjohn Feb 15 '14 at 8:33
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It is the so-called Young's inequality and to prove it you can exploit the concavity of the logarithm: $$ \log (xy) = \frac{1}{p} \log x^p + \frac{1}{q} \log y^q \le \log \left( \frac{1}{p} x^p + \frac{1}{q} y^q \right). $$

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Obviously $p$ and $q$ are not really independent variables - a more natural variable to look at might be $t = 1/p$, so that $1/q = 1-t$. Also make some substitutions $u = x^p$, $v = x^q$. Now you can see that the problem is equivalent to $$\begin{align*} u^{1/p} v^{1/q} &\le \frac{u}{p} + \frac{v}{q} \\ u^{t} v^{1-t} &\le tu + (1-t)v. \end{align*}$$ The presence of the $t$ and $1-t$ should now make it look a lot more like it has to do with convexity.

In fact, if you let $r = u/v$, you can reformulate the problem further by dividing both sides by $v$. This turns the problem into a single variable inequality in $r$ (with a parameter $t$, $0<t<1$): $$r^t \le tr - t + 1.$$ Either of these reformulations should be more approachable than the original problem.

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In addition to this answer, which uses Bernoulli's inequality, we can use Jensen's Inequality and the fact that $e^x$ is convex: $$ e^{x/p+y/q}\le\frac{e^x}{p}+\frac{e^y}{q} $$ where $u=e^{x/p}$ and $v=e^{y/q}$.

Note that since $\frac1p+\frac1q=1$, the measure on two points with weights $\frac1p$ and $\frac1q$ satisfies the requirements of Jensen's Inequality.

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  • $\begingroup$ It was not obvious at first perusal, but now that I look more closely, this is very similar to the approach of Luiz Cordeiro's answer. Since the appearance is fairly different, I will leave this answer as it might be easier to understand by some. $\endgroup$ – robjohn Feb 15 '14 at 8:30
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By AM-GM $$\frac{x^p}{p}+\frac{y^q}{q}\geq\left(x^p\right)^{\frac{1}{p}}\left(y^q\right)^{\frac{1}{q}}=xy$$

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