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From the Wikipedia page on Young's Inequality:

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This above statement of Young's Inequality is the most frequent one that I've encountered in textbooks. Yet consider:

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This second version of Young's Inequality for increasing functions has a very nice, intuitive visual proof (shown above).

Problem: What is connecting Young's Inequality for increasing functions to Young's Inequality for conjugate Holder exponents? In particular, is there a way to prove that

$$ ab \le {a^p \over p} + {b^q \over q} $$

using the proof involving increasing functions, shown visually above?

user1770201
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  • The answers at https://math.stackexchange.com/q/1132153/16490 and https://math.stackexchange.com/q/2253569/16490 prove this inequality, but none of them seem to use Young's inequality to do so... – ziggurism Mar 05 '18 at 14:06
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    Yes, the function $f(x) = x^{p-1}$ is precisely an example of a continuous increasing function with $f(0)=0$, and its inverse is perhaps amusingly $f^{-1}(x)=x^{q-1}$. – Calvin Khor Mar 05 '18 at 14:22
  • Related : https://math.stackexchange.com/questions/149901/geometric-interpretation-of-youngs-inequality?rq=1 – Arnaud D. Mar 05 '18 at 14:26
  • exponent calculation: $1/p + 1/q = 1$ so $q+p=qp$ so $p=q(p-1)$ and $q=p(q-1)$, so that $$ \frac1{p-1} = \frac qp = q-1$$ – Calvin Khor Mar 05 '18 at 14:26
  • @ziggurism sorry it was Arnaud's link that has the picture – Calvin Khor Mar 05 '18 at 14:33
  • Interesting. So this inequality is a generalization of Young's inequality, not an equivalent statement. – ziggurism Mar 05 '18 at 16:04

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