$\blacksquare~$ Problem: Let $f : [0, r) \rightarrow [0, \infty)$ and $f \in \mathscr{C}^{0}$ and strictly increasing with $f(0) = 0$ (the case of $r = \infty$ is allowed). Show that for every $a$ in $[0, r)$ and every $b$ in image of $[0, r)$ under $f$, we have \begin{align*} ab \leqslant \int_{0}^{a} f(x) dx + \int_{0}^{b} f^{-1}(t) dt \end{align*}
$\blacksquare~$ My Approach: We are given a function $f$ such that, \begin{align*} f~:~[ 0 , r ) \rightarrow [ 0 , \infty ) \end{align*} And also given $ f \in \mathscr{C}^{0} $ and strictly increasing with $ f ( 0 ) = 0$.
We also see that, $ f^{- 1} \in \mathscr{C}^0 $ and strictly increasing with $ f^{- 1} ( 0 ) = 0$ .
Now let us substitute $ f^{ - 1 } ( x ) $ by $u$ and obtain the following \begin{align*} & f ( u ) = x\\ \implies & f ' ( u ) du = dx \end{align*}
Therefore we obtain-
- $x \quad \quad \quad~~ 0 \quad \quad b $
- $f^{-1}(x) \quad 0 \quad \quad f^{- 1}(b) $
Now, let's observe that
- $f$ is strictly increasing and continuous on $[0, r)$ for any $r > 0$. Then $f$ is one-one.
Again, we have on substitution- \begin{align*} & \int_{0}^{a} f ( x ) dx + \int_{0}^{f^{- 1} ( b )} u f' ( u ) du\\ = ~ & \int_{0}^{a} f ( x ) dx + b f^{- 1} ( b ) + \int_{f^{- 1} ( b )}^{0} f( u ) du\\ = ~ & \int_{f^{- 1} ( b )}^{a} f ( x ) dx + b f^{- 1} ( b ) \\ \geqslant ~ & b f^{- 1} ( b )\\ \geqslant ~ & ba \end{align*} As we have $ f ( a ) \leqslant b$ $\implies$ $a \leqslant f^{- 1} ( b ) ~$ [As $f^{- 1} $ is increasing ]
Hence we obtain our needed inequality.
$\S~$ Target: We'll basically prove Young's Inequality from here. I have proved it earlier in a different way
$\bullet~$ Young's Inequality: For $x, y > 0$ and $p, q > 1$ such that $\frac{1}{p} + \frac{1}{q} = 1$, then show that \begin{align*} xy \leqslant \frac{x^{p}}{p} + \frac{y^{q}}{q} \end{align*}
$\bullet~$ Extension from Last Problem: Given in problem, \begin{align*} \frac{1}{p} + \frac{1}{q} = 1 \quad \text{Where } p , q \in ( 1 , \infty ) \end{align*} Let us consider the function $ f $ defined as- \begin{align*} f~:~ &[0 , \infty ) \rightarrow [0 , \infty )\\ & x \mapsto x^{p - 1} \quad~ \text{ where } x \in \mathbb{R}^{+} \end{align*} Such that the mapping is defined as following \begin{align*} f ( x ) = x^{p - 1} \quad \text{for any } x \in \mathbb{R}^{+} \end{align*} Now observe that, $ f ( x ) $ is continuous, and strictly increasing in $(0 , \infty )$ as, \begin{align*} f'(x) &= (p - 1)\cdot x^{p - 2} > 0 \quad \text{for any } x \in \mathbb{R}^{+} \end{align*} Now from our given fact about $ p ,q $ we obtain, \begin{align*} &\frac{1}{p} + \frac{1}{q} = 1\\ \implies & \frac{1}{p} = \frac{q - 1}{ q }\\ \implies &(p -1) = \frac{1}{(q - 1) } \end{align*} Therefore, we obtain the inverse mapping for the function given in question as- \begin{align*} f^{- 1} ( x ) & = x^{\frac{1}{( p - 1 )}} \\ &= x^{( q - 1 )} \end{align*} Therefore our function $f$ satisfies all the properties of the previous problem's function. Now from previous problem we have, \begin{align*} & ab ~\leqslant~ \int_{0}^{a} f ( x ) dx + \int_{0}^{b} f^{- 1} ( y ) dy \\ \Leftrightarrow ~ & ab ~\leqslant~ \int_{0}^{a} x^{( p - 1 )} dx + \int_{0}^{b} y^{( q - 1 )} dy \\ \Leftrightarrow ~ & ab ~\leqslant~ \frac{a^p}{p} + \frac{b^{q}}{q} \end{align*} Which gives us our needed inequality.
Please check the proofs and claims for glitches and please add new ideas too :)