Putnam 2003 A2

Let $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ be non-negative real numbers. Show that $$(a_1\ldots a_n)^{1/n} + (b_1\ldots b_n)^{1/n} \le [(a_1 + b_1) \cdots (a_n + b_n)]^{1/n}.$$

Solution: we will use the generalized Holder’s inequality which states that
$$||f_1\ldots f_n ||_1 \le ||f_1||_{\lambda_1} \cdots ||f_n||_{\lambda_n}$$
for lambda weights $\lambda_1^{-1} + \cdots + \lambda_n^{-1} = 1$ all greater than 1.

Assuming this is true, let $f_i = (a_i^{1/n}, b_i^{1/n})$ and the norms be the discrete $l^p$ norm. This will give us $||f_1 \ldots f_n||_1 = (a_1\ldots a_n)^{1/n} + (b_1\ldots b_n)^{1/n}$ as everything is non-negative. The weight will be uniform $\lambda_i = n$, then the right hand side will be
$$||f_i||_{n} = (a_i + b_i)^{1/n}$$
and we have our inequality.

The sole remaining thing to prove is the generalized Holder’s inequality. We will assume the famous base case of the two element case. In the inductive case, we have
\begin{align*}
||f_1\cdots f_{n+1}||_1 &\le ||f_1 \cdots f_n||_{\lambda_{n+1}/(\lambda_{n+1} – 1)} ||f_{n+1}||_{\lambda_{n+1}} \\
&= ||(f_1 \cdots f_n)^{\lambda_{n+1}/(\lambda_{n+1} – 1)}||_1^{(\lambda_{n+1} – 1)/\lambda_{n+1}} ||f_{n+1}||_{\lambda_{n+1}}.
\end{align*}
From here, just change the weights and use the inductive case and we are done.

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