Proving that $\frac{f(b)-f(a)}{b-a}+ \left(\frac{g(b)-g(a)}{b-a}\right)^2\le \max_{t\in [a,b]}\{f'(t)+(g'(t))^2\}$

Question

Let $f,g\in C^1([a,b])$ with $a<b$ then prove that

$$\frac{f(b)-f(a)}{b-a}+ \left(\frac{g(b)-g(a)}{b-a}\right)^2\le \max_{t\in [a,b]}\{f'(t)+(g'(t))^2\}$$

It smells like there is some mean value theorem going around. But I tried it as follows:

Indeed, it springs from mean value theorem that There exists $c_1,c_2\in (a,b)$ such that

$$\frac{f(b)-f(a)}{b-a} = f'(c_1)~~~ and ~~~~\frac{g(b)-g(a)}{b-a} = g'(c_2)$$

Then I have $$\frac{f(b)-f(a)}{b-a}+ \left(\frac{g(b)-g(a)}{b-a}\right)^2= f'(c_1)+(g'(c_2))^2\le \max_{t\in [a,b]}\{f'(t)\}+\max_{t\in [a,b]}\{(g'(t)^2)\}$$

Which is however not the required inequality.

Can anyone help? how can I improve this ?

Answer 1

\begin{align*} &\dfrac{f(b)-f(a)}{b-a}+\left(\dfrac{g(b)-g(a)}{b-a}\right)^{2}\\ &=\int_{a}^{b}f'(t)\dfrac{dt}{b-a}+\left(\int_{a}^{b}g'(t)\dfrac{dt}{b-a}\right)^{2}\\ &\leq\int_{a}^{b}f'(t)\dfrac{dt}{b-a}+\int_{a}^{b}(g'(t))^{2}\dfrac{dt}{b-a}\\ &\leq\max_{t\in[a,b]}\{f'(t)+(g'(t))^{2}\}\int_{a}^{b}\dfrac{dt}{b-a}\\ &=\max_{t\in[a,b]}\{f'(t)+(g'(t))^{2}\}. \end{align*}

Answer 2

Treat $\max$ as $\sup$, then apply then fact that $\sup(A+B)=\sup(A)+\sup(B)$, for $A,B$ be nonempty bounded subset of $\mathbb R$ and $A+B=\{a+b:a\in A,b\in B\}$