Proving Young's inequality for n numbers whitout using AM-GM inequality

Question

Let $\alpha_1, ..., \alpha_n \geq 0$, and $p_1,...,p_n \geq 0$ such that $\sum_{i=1}^{n} p_i^{-1} =1$. Show that $\alpha_1...\alpha_n \leq p_{1}^{-1}\alpha_1^{p_1}+...+p_{n}^{-1}\alpha_n^{p_n}$.

It is Young's inequality for n numbers. I know the proof that uses AM-GM, but in this case, I can't use it.

My attempt: I tried to solve it by induction, I suppose that it was true for $n-1$ numbers and summed all the possibilities to obtain $\sum_{k=1}^{n} \prod_{i \in ([k]-\{k\})} \alpha_i \leq (n-1)(p_{1}^{-1}\alpha_1^{p_1}+...+p_{n}^{-1}\alpha_n^{p_n})$, but don't seem to be that the left right hand is greater or equal to $\alpha_1...\alpha_n$.

Answer 1

The usual way is by Jensen's, but you can do this by induction without Jensen's or AM-GM if you want as follows:

Two variables is obvious from non AM-GM means, e.g., consider areas with the curve $x_2=x_1^{p_1-1}$ (which is the same as $x_1=x_2^{p_2-1}$) on nonnegative reals.

Then inductively, suppose we have prove the $n$-variable case for all $n\leq N$: all $p=(p_1,\dots,p_n)$ and all $\alpha=(\alpha_1,\dots,\alpha_n)$ satisfying the condition we have Young's inequality holds. To prove the case for the $N+1$ variable case $p=(p_1,\dots,p_N,p_{N+1})$ and $\alpha=(\alpha_1,\dots,\alpha_{N+1})$ (so $N\geq 2$), first use the two-variable case on $\tilde{p}=(p_N/C, p_{N+1}/C)$ and $\tilde{\alpha}=(\alpha_N^C,\alpha_{N+1}^C)$ (where the constant $C=1/(p_N^{-1}+p_{N+1}^{-1})$ is chosen to make sure the hypothesis is satisfied): $$ \alpha_N^C\alpha_{N+1}^C\leq Cp_N^{-1}\alpha_N^{p_N}+Cp_{N+1}^{-1}\alpha_{N+1}^{p_{N+1}} $$ and so with the $N$-variable case $\alpha'=(\alpha_1,\dots,\alpha_{N-1},\alpha_N\alpha_{N+1})$ and $p'=(p_1,\dots,p_{N-1},C)$: \begin{align*}\require{color} \alpha_1\dots\alpha_{N-1}\alpha_N\alpha_{N+1} &\leq p_1^{-1}\alpha_1^{p_1}+\dots+p_{N-1}^{-1}\alpha_{N-1}^{p_{N-1}}+C^{-1}(\alpha_N\alpha_{N+1})^C\\ &\leq p_1^{-1}\alpha_1^{p_1}+\dots+p_{N-1}^{-1}\alpha_{N-1}^{p_{N-1}}+C^{-1}\color{red}{(Cp_N^{-1}\alpha_N^{p_N}+Cp_{N+1}^{-1}\alpha_{N+1}^{p_{N+1}})}\\ &=p_1^{-1}\alpha_1^{p_1}+\dots+p_{N-1}^{-1}\alpha_{N-1}^{p_{N-1}}+p_N^{-1}\alpha_N^{p_N}+p_{N+1}^{-1}\alpha_{N+1}^{p_{N+1}} \end{align*} as desired.

Answer 2

We note that Jensen's Inequality states that for a convex function $f:I\to\mathbb{R}$ defined on an interval $I$, given $x_1,x_2,\dots,x_n\in I$ and $\omega_1,\omega_2,\dots,\omega_n\geq 0$, then we have $$f\left(\frac{\omega_1x_1+\omega_2x_2+\dots+\omega_nx_n}{\omega_1+\omega_2+\dots+\omega_n}\right)\leq\frac{\omega_1f(x_1)+\omega_2f(x_2)+\dots+\omega_nf(x_n)}{\omega_1+\omega_2+\dots+\omega_n} $$

or more compactly, $$f\left(\frac{\sum_{i=1}^{n}\omega_ix_i}{\sum_{i=1}^{n}\omega_i}\right)\leq\frac{\sum_{i=1}^{n}\omega_if(x_i)}{\sum_{i=1}^{n}\omega_i} $$

Let $\gamma_1,\gamma_2,\dots,\gamma_n$ be positive real numbers and let $p_1,p_2,\dots,p_n$ be positive real numbers such that $\sum_{i=1}^{n}p_i^{-1}=1$.

Recall that the function $\exp:\mathbb{R}\to(0,\infty)$ is convex, since its second derivative is always positive.

Now, we observe that we can write our weights $\omega_i$ as $p_i^{-1}$ and our $x_i$ as $\ln(\gamma_i^{p_i})$. Then we have

$$\begin{align*}\prod_{i=1}^{n}\gamma_i&=\exp\left(\ln\left(\prod_{i=1}^{n}\gamma_i\right)\right)\\ &=\exp\left(\sum_{i=1}^{n}\ln(\gamma_i)\right)\\ &=\exp\left(\sum_{i=1}^{n}p_i^{-1}\ln(\gamma_i^{p_i})\right)\tag{Mult. by $1=p_i\cdot p_i^{-1}$}\\ &=\exp\left(\frac{\sum_{i=1}^{n}p_i^{-1}\ln(\gamma_i^{p_i})}{\sum_{i=1}^{n}p_i^{-1}}\right)\tag{Divide by $1$}\\ &\leq\frac{\sum_{i=1}^{n}p_i^{-1}\exp(\ln(\gamma_i^{p_i}))}{\sum_{i=1}^{n}p_i^{-1}}\tag{Applying Jensen's}\\ &=\sum_{i=1}^{n}p_i^{-1}\gamma_i^{p_i}\tag{$\sum_{i=1}^{n}p_i^{-1}=1$}\end{align*}$$