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I am trying to prove Young's Inequality by considering the function $$h(u) = \frac{u^p}{p} + \frac{C^q}{qu^q}$$ for $C,u>0$ and $p,q >1$. We also require $$\frac{1}{p}+\frac{1}{q}=1$$ so that $pq=p+q$. I starting by using straightforward calculus to show that $h$ has a global minimum at $u=C^{\frac{1}{p}}$.

Next, I would like to show that $C \leqslant h(u)$ for any $C,u>0$. If I can do this, I will be able to let $C=uv$ for $u,v\geqslant 0$ and arrive at Young's Inequality. However, I am having trouble doing so. I feel like there is something very simple that I'm just not seeing in connecting these steps. Any help would be greatly appreciated. Thanks.

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  • $\begingroup$ If you have found the minimum using calculus, then $h(u)\ge h(u_{min})=h(C^{1/p})$ gives you the result. Alternately, weighted AM-GM does the same. $\endgroup$
    – Macavity
    Jan 28, 2015 at 17:22
  • $\begingroup$ @Macavity But isn't it not necessarily the case that $C < C^{1/p}$? So how can we say that $C\leq h(u)$? $\endgroup$
    – Ray B.
    Jan 28, 2015 at 17:30
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    $\begingroup$ What does it mean to say that the global minimum of $h(u)$ is when $u=C^{\frac1p}$? $\endgroup$
    – Macavity
    Jan 28, 2015 at 18:24
  • $\begingroup$ Then note that $h(C^{\frac1p})=C$. $\endgroup$
    – Macavity
    Jan 28, 2015 at 18:26
  • $\begingroup$ @Macavity Ah yes, I see. It turns out I only showed that $C^{1/p}$ was a local minimum. But indeed it is a global minimum and now the result follows. Thanks. $\endgroup$
    – Ray B.
    Jan 28, 2015 at 19:18

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