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Martin Sleziak
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Understanding the condition $\frac{1}{p} + \frac{1}{q} = 1$ in the estimate $xy \le \frac{1}{p}x^p + \frac{1}{q}y^q$

I just read a proof of Holder's inequality in measure theory, which boils down to the following inequality: $$xy \le \frac{1}{p}x^p + \frac{1}{q}y^q$$ where $x,y\ge 0$ and $\frac{1}{p} + \frac{1}{q} = 1$.

This inequality can be proved by considering the difference, viewed as a function in $x$, and showing, using calculus, that its minimum is 0.

My question is: Why the heck should this be true? What is the meaning behind the condition that $\frac{1}{p} + \frac{1}{q} = 1$? To me, this reminds me of the theory of triangle groups, where the properties of the triangle group $\Delta(p,q,r) := \langle a,b,c|a^p = b^q = c^r = abc = 1\rangle$ is very sensitive to whether $\frac{1}{p} + \frac{1}{q} + \frac{1}{r}$ is less than, greater than, or equal to 1. I don't know if there's any resemblance?