TITLE: Proving Holder's inequality using Jensen's inequality
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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.)