I am given a relation that for $a,b>0$ we have $$ab\leq \dfrac{a^p}{p}+\dfrac{b^q}{q}$$ where $ p$ and $ q$ satisfy $\dfrac{1}{p}+\dfrac{1}{q}=1$. Now how do I use this to prove the AM-GM inequality?
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1What if $p=q=2$? – Botond Jul 23 '19 at 20:26
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then $2ab\leq a^2+b^2\leq (a+b)^2$ – Upstart Jul 23 '19 at 20:27
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Yes, $ab \leqslant \frac{a^2+b^2}{2}$. Isn't it familiar? – Botond Jul 23 '19 at 20:28
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okay so we use that $\sqrt{ab}\leq ab$ – Upstart Jul 23 '19 at 20:31
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Not exactly. We have that $\forall a,b > 0$, $ab \leqslant \frac{a^2+b^2}{2}$. And we want to prove that $\forall x, y > 0$, $\sqrt{xy} \leqslant \frac{x+y}{2}$. Aren't they similar? – Botond Jul 23 '19 at 20:32
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ohhhh...since it is true for all $x,y$ then it is true for $\sqrt{x}$ and$ \sqrt{y}$ also – Upstart Jul 23 '19 at 20:34
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1Yes! If we let $a=\sqrt{x}$ and $b=\sqrt{y}$, then we have that $\forall x,y > 0$, $\sqrt{xy} \leqslant \frac{x+y}{2}$. And we can do it, because $x \to \sqrt{x}$ is a bijection between $(0,\infty)$ and $(0,\infty)$. – Botond Jul 23 '19 at 20:40
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If you introduce $x=a^p$, $y=b^q$ $s=1/p$ and $t=1/q$, then this is the weighted AM-GM: $$ x^sy^t\leq sx+ty $$ You can deduce this from the regular AM-GM for rational $s$ and $t$, but for real $s$ and $t$ you need some additional argument, like continuity.
Arthur
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