# Prove $ab\leq F(a)+G(b),~~\text{for all}~a\geq 0$ and $b\geq 0$

Suppose that te function $f:[0,\infty)\rightarrow\mathbb{R}$ is continuous and strictly increasing, with $f(0)=0$ and $f([0,\infty))=[0,\infty)$. Then define $$F(x)=\int_{0}^{x}f\qquad\text{and}\qquad G(x)=\int_{0}^{x}f^{-1}\qquad\text{for all x\geq 0}$$ Prove Young's Inequality: $$ab\leq F(a)+G(b)\qquad\text{for all a\geq0 and b\geq0}$$

For this question, I think I need to use this formula:

I tried to use Cauchy Inequality, but it didn't give me any clews keep going to solve the problem. Can someone give me a hint or suggestion to start? Thanks.

Consider the function $\phi(b)=ab-\int_0^af-\int_0^bf^{-1}$. Then: $$\phi'(b)=a-f^{-1}(b)$$ Therefore $\phi(b)$ has an extremum at $b=f(a)$ where $\phi(b)=0$ using the formula you mentioned in the question. Also $\phi''(b) = -f^{-1'}(b)<0$ since $g'(f(x))f'(x)=1$ and $f'(x)>0$ (strictly increasing) where $g=f^{-1}$. Therefore, this function $\phi(b)$ has a maximum value of $0 \implies \phi(b)\leq0$ which is simply the inequality to be proved.
Case 1, $f^{-1}(b) \ge a$. Then \begin{align*} \int_0^a f(u)\, du + \int_0^b f^{-1}(u)\, du &=\int_0^a f(u)\, du+\int_0^{f^{-1}(b)} u\, df(u)\\ &=\int_0^a f(u)\, du+\int_0^a u\, df(u) + \int_a^{f^{-1}(b)} u\, df(u)\\ &\ge \int_0^a f(u)\, du+\int_0^a u\, df(u) + \int_a^{f^{-1}(b)} a\, df(u)\\ &= af(a) + a \big[b -f(a) \big]\\ &= ab. \end{align*}
Case 2, $f^{-1}(b) \le a$. Then \begin{align*} \int_0^a f(u)\, du + \int_0^b f^{-1}(u)\, du &=\int_0^a f(u)\, du+\int_0^{f^{-1}(b)} u\, df(u)\\ &=\int_0^{f^{-1}(b)} f(u)\, du+\int_0^{f^{-1}(b)} u\, df(u) + \int_{f^{-1}(b)}^a f(u)\, du\\ &\ge \int_0^{f^{-1}(b)} f(u)\, du+\int_0^{f^{-1}(b)} u\, df(u) + \int_{f^{-1}(b)}^a b\, du\\ &= f^{-1}(b) b + b \Big[a -f^{-1}(b) \Big]\\ &= ab. \end{align*}