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I found a very informal proof (geometric one, for by taking $\alpha$ as $\beta$ in the x-axis and y-axis and showing rectangle area is smaller than area under the plots). Is there a proper algebraic proof for the inequality ?

$$ \alpha\beta \leq \frac{\alpha^p}{p} + \frac{\beta^q}{q} $$ where $\alpha, \beta > 0$ $$ \frac{1}{p} + \frac{1}{q} = 1 $$

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  • $\begingroup$ Would taking integrals count as an algebraic proof? I think that it will be very close to the geometric proof... $\endgroup$ – Shashi Jan 10 '18 at 19:47
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    $\begingroup$ Just search for Young's inequality. There is also a proof on the Wikipedia page. $\endgroup$ – Fritz Jan 10 '18 at 19:47
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For positives our $p$ and $q$ it's just AM-GM: $$\frac{1}{p}\alpha^p+\frac{1}{q}\beta^q\geq\left(\alpha^p\right)^{\frac{1}{p}}\left(\beta^q\right)^{\frac{1}{q}}=\alpha\beta$$

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You might use the convexity of $f:\mathbb{R}\rightarrow \mathbb{R}^+,t\mapsto \exp(t)$: $$\alpha\beta=\exp(\log(\alpha)+\log(\beta))=\exp(\frac{1}{p}\log(\alpha^p)+\frac{1}{q}\log(\alpha^q))\underbrace{\leq}_{convex. of \exp}$$$$\frac{1}{p}\exp(\log(\alpha^p))+\frac{1}{q}\exp(\log(\beta^q))=\frac{1}{p}\alpha^p+\frac{1}{q}\beta^q$$ which is not really proper algebraic, but hopefully clearer and more rigorous than the proof You read.

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