Proving Holder's inequality using Jensen's inequality Let $p$ and $q$ be positive reals such that $\frac{1}{p}+\frac{1}{q} = 1$, so that $p,q$ in $(1,\infty)$.
For $\vec a$ and $\vec b \in \mathbb{R}^2$ prove that $|\vec a \cdot \vec b | \leq ||\vec a||_p|| \vec b||_q$.
A hint was posted for using Jensen's inequality to use $\phi(x) = ln(1 + e^x)$. But I don't know how I'd work that in.
 A: It is easy to get from Jensen to Young to Holder.  However if you really want to do directly, note it is sufficient to show:
$$ \sum_{k=1}^n \lvert a_k \rvert \lvert b_k \rvert \le \left(\sum_{k=1}^n \lvert a_k \rvert^p \right)^{\frac1p}\left(\sum_{k=1}^n \lvert b_k \rvert^q \right)^{\frac1q} \tag{1}$$
for $\lvert a_k \rvert > 0$ (why?).
As $x^q$ is convex in $(0, \infty)$, by Jensen inequality we have $\displaystyle \left(\sum_{k=1}^n w_k x_k\right)^q \le \sum_{k=1}^n w_k x_k^q$ for $x_k, w_k >0$ and $\sum_k w_k = 1$.
Using $w_k = \dfrac{|a_k|^p}{\sum_k |a_k|^p}$ and $x_k = \dfrac{|a_k||b_k|}{w_k}$ in the above form of Jensen Inequality, we can get $(1)$.
A: We want to show that
$$ a_1b_1 + a_2b_2 \le (a_1^p+a_2^p)^{1/p} (b_1^q+b_2^q)^{1/q} $$
First note that the inequality is homogeneous in both $(a_1,a_2)$ and $(b_1,b_2)$, separately.  Thus we can scale them both to halve the number of variables involved: dividing both sides by $a_1b_1$, we get
$$ 1 + \frac{a_2}{a_1}\cdot\frac{b_2}{b_1}
\le \Bigl(1+\Bigl(\frac{a_2}{a_1}\Bigr)^p\Bigr)^{1/p}
\Bigl(1+\Bigl(\frac{b_2}{b_1}\Bigr)^q\Bigr)^{1/q} $$
Write $u=a_2/a_1$ and $v=b_2/b_1$; we then want to show
$$ 1 + uv \le (1+u^p)^{1/p} (1+v^q)^{1/q} $$
Taking logarithms, we get the equivalent
$$ \ln(1+uv) \le \frac1p \ln(1+u^p) + \frac1q \ln (1+v^q) $$
Writing $x=p\ln u$ and $y=q\ln v$, this is equivalent to
$$ \ln(1+e^{x/p+y/q}) \le \frac1p \ln(1+e^x) + \frac1q \ln (1+e^y) $$
which asserts the convexity of $x\mapsto\ln(1+e^x)$.  So, check the second derivative and you're done.
(This argument generalizes to give the inequality in this question, which was question A2 on the 2003 Putnam; see Kedlaya's archive for that solution and some other nice ones.)
A: Here's first proving an easier version: Note $\phi(x) = -\log x $ is convex, on $x > 0,$ and hence convexity (= Jensen) yields
$$
-\log(tx + (1-t)y) \leq -t\log x - (1-t)\log y,
$$
let $x = u^p, y = v^q,$ and $t = 1/p,$ where $u,v > 0.$  You then easily get,
$$
uv \leq \frac{u^p}{p} + \frac{v^q}{q}.
$$
Now, if $\|a\|_p = \|b\|_q = 1,$ then we see 
$$
|\sum_{i=1}^n a_i b_i| \leq \sum_{i=1}^n |a_i||b_i| \leq \sum_{i=1}^n \frac{|a_i|^p}{p} + \sum_{i=1}^n \frac{|b_i|^q}{q} = 1. 
$$
For general vectors, just normalize. 
