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 $a<b$ 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 ?

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 $A+B=\{a+b:a\in A,b\in B\}$
