Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $x, y, z \geq 0$ 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 $$ xyz \leq \frac{x^p}{p} + \frac{y^q}{q} + \frac{z^r}{r} $$ 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 $$ xy \leq \frac{x^p}{p} + \frac{y^q}{q} $$ 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.

share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer
Very nice, thank you ! I did not realize I could just apply it to $y^{\frac{p}{p-1}}$ and $z^{\frac{p}{p-1}}$... – Amateur May 30 '13 at 17:27

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 $x^p = y^q = z^r$.

share|cite|improve this answer
Thanks for the answer. I think you meant $\ln$ is strictly concave (not strictly convex). However, I would like to apply the two-parameters Young's inequality twice instead of using Jensen's inequality. – Amateur May 30 '13 at 16:50
@Frank: Thanks for catching the typo. – copper.hat May 30 '13 at 16:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.