Young's inequality for three variables Let $x, y, z \geqslant 0$ and let $p, q, r > 1$ be such that 
$$
\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 1.
$$
How can one show that under these hypotheses we have
$$
xyz \leqslant \frac{x^p}{p} + \frac{y^q}{q} + \frac{z^r}{r}
$$
with equality if and only if $x^p = y^q = z^r$, using twice the standard two-parameters Young's inequality which says that for all $x, y \geq 0$ and for all $p, q > 1$ for which $\frac{1}{p} + \frac{1}{q} = 1$ we have
$$
xy \leqslant \frac{x^p}{p} + \frac{y^q}{q}
$$
with equality if and only if $x^p = y^q$ ?
I've tried to apply it twice directly, to multiply two inequalities and to add two inequalities, but in each case it gets quite messy and I can't get the desired result, even though I'm sure it should be quite simple.
 A: The function $\ln$ is concave, so if $\sum_n \lambda_n =1$, with $\lambda_n \ge 0$, then $\ln ( \sum_n \lambda_n x_n ) \ge \sum_n \lambda_n \ln x_n$ (with $x_n >0$, of course).
Hence $\ln ( \frac{x^p}{p} + \frac{y^q}{q} + \frac{z^r}{r} ) \ge \frac{1}{p} \ln x^p + \frac{1}{q} \ln y^q + \frac{1}{r} \ln x^r = \ln (x y z)$.
Taking exponents yields the desired result.
Since $\ln$ is strictly concave, we have equality iff $x_i = x_j$ in the first inequality, which corresponds to $x^p = y^q = z^r$.
A: Here's a solution by applying Young's inequality twice.
First apply the inequality to $x$ and $yz$ with $p$ and $\frac{p}{p-1}$ to get
$$xyz \le \frac{x^p}{p} + \frac{(yz)^{\frac{p}{p-1}}}{\frac{p}{p-1}}.$$
Then apply it to $y^{\frac{p}{p-1}}$ and $z^{\frac{p}{p-1}}$ with $\frac{p-1}{p} q$ and $\frac{\frac{p-1}{p}q}{\frac{p-1}{p}q - 1} = \frac{(p-1)q}{pq-p-q}$ to get
$$xyz \le \frac{x^p}{p} + \frac{p-1}{p} \left(\frac{y^q}{\frac{p-1}{p}q} + \frac{z^{\frac{pq}{pq-p-q}}}{\frac{(p-1)q}{pq-p-q}}\right)$$
Notice that $\frac{pq}{pq-p-q} = r$, so you get
$$xyz \le \frac{x^p}{p} + \frac{y^q}{q} + \frac{z^r}{r}$$
as wanted.
