Inequality $\frac{{b{e^{bx}} - a{e^{ax}}}}{{b - a}}How to prove $$\displaystyle \frac{{b{e^{bx}} - a{e^{ax}}}}{{b - a}}<e^{(a+b)x}$$ for $0<a<b$ and $x\ne 0$ ?
 A: Hint: Use Jensen inequality for $e^x$ with suitable weights...
A: The proof independent of Jensen's: multiply both sides by the denominators and group the terms to end up with
$$\frac{be^{bx}}{e^{bx}-1}\ge \frac{ae^{ax}}{e^{ax}-1}.$$
Now fix $x$ and consider the function $f(t)=\frac{te^{tx}}{e^{tx}-1}=t+\frac{t}{e^{tx}-1}.$ Note, $f'(t)=\frac{e^{tx}(e^{tx}-1-tx)}{(e^{tx}-1)^2}\ge 0,$ since $e^y\ge 1+y$ and thus $f(b)\ge f(a)$ for $b\ge a>0.$
A: As mentioned in the comments, we can rearrange the original inequalities as follows:
$$
\displaystyle {{b{e^{bx}} }}<(b-a)e^{(a+b)x}+a{e^{ax}}
$$
Here you can use Jensen but if you do not want to use notion of convexity, you can use the weighted arithmetic geometric inequality, which is:
$$
w_1x_1+w_2x_2\geq (w_1+w_2)(x_1^{w_1}x_2^{w_2})^{\frac{1}{w_1+w_2}}
$$
which gives you:
$$
\displaystyle (b-a)e^{(a+b)x}+a{e^{ax}}\geq  (b-a+a)(e^{a.ax+(b-a)(b+a)x})^{\frac{1}{b}}
$$

This does NOT mean that the proof is independent of Jensen inequality. It is only so, if you do not  use Jensen inequality to prove weighted AG inequality.
