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?
It is the so-called Young's inequality and to prove it you can exploit the concavity of the logarithm: $$ \log (xy) = \frac{1}{p} \log x^p + \frac{1}{q} \log y^q \le \log \left( \frac{1}{p} x^p + \frac{1}{q} y^q \right). $$
The exponential funtion $t\mapsto \exp(t)$ is convex, so $$\begin{align} xy&=\exp(\log(xy))\\ &=\exp(\log(x)+\log(y))\\ &=\exp((1/p)\log(x^p)+(1/q)\log(y^q))\\ &\leq (1/p)\exp(\log(x^p))+(1/q)\exp(\log(y^q))\\ &=\frac{x^p}{p}+\frac{y^q}{q}\\ \end{align}$$
By AM-GM $$\frac{x^p}{p}+\frac{y^q}{q}\geq\left(x^p\right)^{\frac{1}{p}}\left(y^q\right)^{\frac{1}{q}}=xy$$
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.
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.
