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Let $p>1$ and put $q=\frac{p}{p-1}$, so $1/p+1/q=1$. Show that for any $x>0$ and $y>0$, we have $$ xy \le \frac{x^p}{p}+\frac{y^q}{q}$$ And find where the equality holds.

So far, I have simply tried to multiply through the RHS of the above expression and see what would happen, plugged in for $q$ and I got this:

$$ pxy \le x^p+(p-1)y^\frac{p}{p-1} $$ We also know that $q>1$ by its definition and using $p>1$, but I am not quite sure how to proceed. Any suggestions? Thank you for the help

johnsteck
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1 Answers1

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Hint: Since $x > 0$, write $$x = e^{t/p}$$ for some $t$. Likewise $y = e^{s/q}$. Then

$$xy = e^{t/p + s/q} \le \frac{e^t}{p} + \frac{e^s}{q}$$

since the exponential is convex.

  • Thanks! I completely forgot about using the properties of convexity. So then the case $p=1$ and $q=1$ would satisfy the equality if I am not mistaken. – johnsteck Oct 17 '13 at 06:20
  • @Darkel90 You've assumed that $p > 1$, though. Note that $e^z$ is strictly convex, though, so... –  Oct 17 '13 at 06:26
  • So would there not be a solution for $p$ and $q% such that the equality holds? – johnsteck Oct 17 '13 at 06:29
  • @Darkel90 Equality holds if and only if $x^p = y^q$ iff $e^t = e^s$ iff $s = t$. –  Oct 17 '13 at 06:34
  • Thank you. It's been a long day and I really should have seen this last part. I appreciate your help :) – johnsteck Oct 17 '13 at 06:38
  • @Darkel90 You're very welcome :) –  Oct 17 '13 at 06:38