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The inequality is the following:

$a^ \theta b^ {1-\theta}$ $\leq$ $[\theta ^ \theta (1-\theta)^ {1-\theta}]^{1/p}(a^p+b^p)^{1/p}$, where $\theta \in [0,1]$, $a,b$ are nonnegative.

This inequality is used to give a sharper constant in the proof of an embedding theorem in Sobolev spaces. Here is the link https://www.math.ucdavis.edu/~hunter/pdes/ch3.pdf. On page 66, the author used the inequality to give a sharper estimate, but he didn't give a proof of this inequality. Actually, $a^ \theta b^ {1-\theta}$ $\leq$ $(a^p+b^p)^{1/p}$ is obvious (this is my first try), and enough to prove the embedding theorem, but it is always interesting to give a sharper inequality.

I tried to prove this seemingly elementary inequality, but I'm really not good at it. Can anyone give a smart answer? Any hint would be appreciated.

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3 Answers 3

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Both sides of the inequality are homogeneous, if we multiply $a$ and $b$ with $t > 0$, the expression is multiplied with $t$. Therefore, we can assume $a^p + b^p = 1$ (if $a = b = 0$, the inequality is trivially an equality). So we need to show

$$ a^\theta b^{1-\theta} \leqslant \left[\theta^\theta(1-\theta)^{1-\theta}\right]^{1/p}$$

under the constraint $a^p+b^p = 1$.

If $a = 0$ or $b = 0$, the inequality is again trivial, so let's suppose $a \neq 0 \neq b$.

Let $\alpha = a^p$ and raise the inequality to the $p$-th power to obtain

$$\alpha^\theta(1-\alpha)^{1-\theta} \leqslant \theta^\theta(1-\theta)^{1-\theta}$$

as the inequality to be shown. Take the logarithm and differentiate with respect to $\alpha$:

$$\frac{d}{d\alpha} \left(\log \alpha^\theta(1-\alpha)^{1-\theta}\right) = \frac{\theta}{\alpha} - \frac{1-\theta}{1-\alpha}.$$

It is easy to see that the derivative is positive for $0 < \alpha < \theta$ and negative for $\theta < \alpha < 1$, hence $\alpha \mapsto \alpha^\theta(1-\alpha)^{1-\theta}$ has a unique maximum at $\alpha = \theta$.

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  • $\begingroup$ Thanks very much. Even though I realized the homogeneity at first, I didn't think about it in this way. The proof is really smart. $\endgroup$
    – student
    May 16, 2014 at 22:25
  • $\begingroup$ I'll vote for this answer because it provided the idea to deal with this kind of problem, although it is more complicated than the others. $\endgroup$
    – student
    May 16, 2014 at 22:39
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By the AM/GM inequality, $$ \left(\frac a\theta\right)^\theta \left(\frac b{1-\theta}\right)^{1-\theta} \le \theta\left(\frac a\theta\right) + (1-\theta)\left(\frac b{1-\theta}\right) = a+b $$ Now replace $a$ and $b$ with $a^p$ and $b^p$.

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  • $\begingroup$ Thanks very much! This may be the most elegant proof. $\endgroup$
    – student
    May 16, 2014 at 22:37
  • $\begingroup$ It's really the same as the earlier answer by @TenaliRaman. $\endgroup$
    – user21467
    May 16, 2014 at 22:48
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From Jensen, we can obtain, $$a^{\theta}b^{1 - \theta} \leq \theta a + (1 - \theta) b$$ for $a, b \geq 0$ and $0 \leq \theta \leq 1$.

Substitute $a = (1 - \theta)a$ and $b = \theta b$, and we get, $$((1 - \theta)a)^{\theta}(\theta b)^{1 - \theta} \leq \theta (1 - \theta) a + (1 - \theta) \theta b$$ which gives, $$a^{\theta}b^{1 - \theta} \leq \theta^\theta (1 - \theta)^{1 - \theta} (a + b)$$

The rest should follow.

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  • $\begingroup$ Thanks a lot. Jensen's inequality makes the proof simpler. And for the last step, replace $a$ and $b$ with $a^p$ and $b^p$ and take the p-th root then get the desired result. $\endgroup$
    – student
    May 16, 2014 at 22:36
  • $\begingroup$ Precisely. (Surely you meant pth root instead of pth power). $\endgroup$ May 16, 2014 at 22:40
  • $\begingroup$ Oh, thanks for telling me the typo. I've fixed it. $\endgroup$
    – student
    May 16, 2014 at 22:41

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