Let $a_1,\dots,a_d,b_1,\dots,b_d$ be positive real numbers. Then $$ \prod_{i=1}^d a_i+\prod_{i=1}^d b_i \le \prod_{i=1}^d (a_i^d+b_i^d)^{1/d}$$ and equality holds if and only if $\frac{a_i}{a_j}=\frac{b_i}{b_j}$.
Is there a nice proof for this fact? Is there some geometric interpretation?
I know a proof based on an application of Holder's inequality which seems rather miraculous to me. The idea is to define $f_i:\{0,1\} \to \mathbb{R}$, by $f_i(0)=a_i,f_i(1)=b_i$, and to endow the domain $\{0,1\}$ with the counting measure. The generalized Holder's inequality then states that
$$\prod_{i=1}^d a_i+\prod_{i=1}^d b_i\le\| \prod_{i=1}^d f_i\|_1 \leq \prod_{i=1}^d \|f_i\|_d=\prod_{i=1}^d (a_i^d+b_i^d)^{1/d}$$
and equality holds if and only if $\frac{f_i}{f_j}$ is constant, i.e. $\frac{a_i}{a_j}=\frac{b_i}{b_j}$. Somehow, I feel that it's weird to use an "abstract" measure-theoretic inequality, applied to such a degenerate case (finite measure space with the counting measure).
Are there other, "more elementary" proofs?
Even the case where $a_i=c$ are constant does not seem trivial to me: i.e. why
$$ c^d+\prod_{i=1}^d b_i \le \prod_{i=1}^d (c^d+b_i^d)^{1/d}$$ and equality holds if and only if $b_1=b_2=\dots=b_d$.
I tried using the generalized Young's inequality for multiple products, which states that $$ \prod_{i=1}^d a_i \le \frac{\sum_i {a_i^d}}{d}.$$ (This is in fact the AM-GM inequality in disguise.)
This implies
$$ \prod_{i=1}^d a_i+\prod_{i=1}^d b_i \le \frac{\sum_i {a_i^d}+{b_i^d}}{d} \ge \prod_{i=1}^d (a_i^d+b_i^d)^{1/d}$$
but the direction of the last inequality (also by AM-GM) is opposite to the one we would like to have.
Of course, we can always mimic the (inductive) proof of the generalized Holder inequality, but this does not seem enlightening to me.
