Proving $e^x\leq e^a\frac{b-x}{b-a}+e^b\frac{x-a}{b-a}$ I'm trying to prove 
$$e^x\leq e^a\frac{b-x}{b-a}+e^b\frac{x-a}{b-a}$$
for any $x\in[a,b]$.  Since this looks reminiscent of the mean value theorem or linear approximations I jotted down some equations relating to those, but didn't see any way of making progress with them.  I know that $e^x$ is an increasing function so if I could perhaps show that the value on the right is equal to $e$ to some value and prove that value is greater than $x$, it would be sufficient.  But I'm not seeing any way to make that work either.
The right-hand side is also equal to this line
$$\left(\frac{e^b-e^a}{b-a}\right)x+\frac{e^ab-e^ba}{b-a}$$
But I can't think of how I would prove that two curves don't intersect in a region.  

We can consider the function $$g(x) =f(b) - f(x) - \frac{f(b) - (a)} {b-a} (b-x) $$ where $f(x) = e^{x} $. We have to prove that $g(x) \geq 0$ for all $x\in[a, b] $. We have via mean value theorem $$f'(c) =\frac{f(b) - f(a)} {b-a} $$ for some $c\in(a, b) $ and since $f'(c) =f(c) $ we get $$g(x) =f(b) - f(x) - (b-x) f(c)$$ We will show that if $x\in(a, b) $ then $g(x) >0$ and we obviously have $g(a) =g(b) =0$. If $c\leq x<b$ then we can see via mean value theorem that $$g(x) =(b-x) (f'(d)-f(c)) =(b-x) (f(d) - f(c)) $$ for some $d\in(x, b)\subseteq(c, b) $. Clearly $f$ is strictly increasing and we have thus $f(c) <f(d) $ and therefore $g(x) >0$ for all $x\in[c, b) $. 
To handle the case when $x\in(a, c] $ just note that $g(x) $ can also be rewritten as $$g(x) =f(a) - f(x) +\frac{f(b) - f(a)} {b-a} (x-a) =f(a) - f(x) +(x-a) f(c)$$ and the proof can be completed as before.
Note that we have used two properties of $f$ here namely $f'(x) =f(x) $ and $f(x) >0$. If one carefully sees the proof one will realize that all we need here is that $f'$ is strictly increasing which can be ascertained if $f''(x) >0$.
