Proving Holder's inequality using Jensen's inequality

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 $(1)$.

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.)