Proof of Young's inequality

Solution 1:

Assuming $f(a)>b$, we have: $$ \color{red}{\int_{0}^{a} f(x)\,dx} = \mu\left(\{(x,y)\in[0,a]\times[0,f(a)]: 0\leq y\leq f(x)\}\right) $$ and: $$ \color{blue}{\int_{0}^{b} f^{-1}(y)\,dy} = \mu\left(\{(x,y)\in[0,a]\times[0,b]: 0\leq x\leq f^{-1}(y)\}\right) $$ so the sum of the two integrals surely exceeds $\mu\left([0,a]\times[0,b]\right)=ab$.

By the way, it is a lot easier just to draw a picture:

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Solution 2:

Let $C$ be the graph of $v = f(u)$ over the interval $[0,f^{-1}(b)]$. If $f(a) > b$, then $f^{-1}(b) < a$, in which case

\begin{align}\int_0^a f(x)\, dx + \int_0^b f^{-1}(x)\, dx &= \int_0^{f^{-1}(b)} f(x)\, dx + \int_0^b f^{-1}(x)\, dx + \int_{f^{-1}(b)}^a f(x)\, dx\\ &= \int_C u\, dv + v\, du + \int_{f^{-1}(b)}^a f(x)\, dx\\ &= \int_C d(uv) + \int_{f^{-1}(b)}^a f(x)\, dx\\ &= bf^{-1}(b) + \int_{f^{-1}(b)}^a f(x)\, dx\\ &> bf^{-1}(b) + b(a - f^{-1}(b))\\ &= ab \end{align}

Similarly if $f(a) < b$, then $\int_0^a f(x)\, dx + \int_0^b f(x)\, dx > ab$. If $f(a) = b$, then $$\int_0^a f(x)\, dx + \int_0^b f^{-1}(x)\, dx = \int_C u\, dv + v\, du = \int_C d(uv) = af(a) = ab.$$