Proving that $\frac{f(b)-f(a)}{b-a}+ \left(\frac{g(b)-g(a)}{b-a}\right)^2\le \max_{t\in [a,b]}\{f'(t)+(g'(t))^2\}$ Let $f,g\in C^1([a,b])$ with $a&lt;b$ 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 ?
\begin{align*}
&amp;\dfrac{f(b)-f(a)}{b-a}+\left(\dfrac{g(b)-g(a)}{b-a}\right)^{2}\\
&amp;=\int_{a}^{b}f'(t)\dfrac{dt}{b-a}+\left(\int_{a}^{b}g'(t)\dfrac{dt}{b-a}\right)^{2}\\
&amp;\leq\int_{a}^{b}f'(t)\dfrac{dt}{b-a}+\int_{a}^{b}(g'(t))^{2}\dfrac{dt}{b-a}\\
&amp;\leq\max_{t\in[a,b]}\{f'(t)+(g'(t))^{2}\}\int_{a}^{b}\dfrac{dt}{b-a}\\
&amp;=\max_{t\in[a,b]}\{f'(t)+(g'(t))^{2}\}.
\end{align*}
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 $A+B=\{a+b:a\in A,b\in B\}$