Examining an inequality involving exponential functions and hyperbolic cosine Let $a,b$ be real numbers with $0 < a < b$.

Problem: I would like to prove/disprove that $$ \frac{a \cdot 2^x+ b \cdot 2^{-x}}{a+b} \leq \cosh(x \log{2})
$$ is true for all real $x \geq 0$.

Approach:


*

*I defined a function $f: \mathbb{R} \to \mathbb{R}$ with $$ f(x) = \cosh(x \log{2}) - \frac{a \cdot 2^x+ b \cdot 2^{-x}}{a+b}.
$$ In order to show the inequality, it suffices to show $f(x) \geq 0$ for all $x \geq 0$.

*I tried to plot the function for some chosen parameters like $a=1$ and $b=2$. In all those cases the function was non-negative, so I suppose that this inequality is true.

*It is $f(0) = 1-1 = 0 \geq 0$.

*Now $f$ is differentiable, so I computed
$$ f'(x) = \frac{\log{2}}{2(a+b)} \big(a (2^x - 2^{-x} - 2^{x+1}) + b(2^x-2^{-x}+2^{-x+1} ) \big). $$

*I would be done if I could show that $f'(x) \geq 0$ for all $x \geq 0$, so $f$ is monotonically increasing and we get our desired result. However, I can not see how this can be shown.


Could you please help me with this problem? That would be nice, thank you in advance!

Consider the following:
$$2^{x}-2^{-x}-2^{x+1}=2^x-2^{-x}-2\cdot2^x= -2^x-2^{-x}=-2(2^{x-1}+2^{-x-1}) = - 2 \left(\frac{2^x+2^{-x}}{2}\right)  $$
Hence
$$2^{x}-2^{-x}-2^{x+1} = - 2 \left(\frac{2^x+2^{-x}}{2}\right) = -2 \cosh(x \log 2)$$
Similarly
$$2^{x}-2^{-x}+2^{1-x}=2^x-2^{-x}+2\cdot2^{-x}= 2^x+2^{-x}=2(2^{x-1}+2^{-x-1}) =  2 \left(\frac{2^x+2^{-x}}{2}\right)  $$
Hence 
$$2^{x}-2^{-x}+2^{1-x}=  2 \left(\frac{2^x+2^{-x}}{2}\right)  = 2 \cosh(x\log 2)$$
So with this the derivative is:
\begin{align}
f'(x)&= \frac{\log 2}{2 (a+b)} \left[-2a \cosh(x \log 2) + 2b \cosh(x \log 2)\right] \\
&= \frac{\log 2}{ a+b} \left[-a \cosh(x \log 2) + b \cosh(x \log 2)\right]\\
&=\log 2\frac{b-a}{a+b} \cosh(x\log2)
\end{align}
with $\log 2 >0$, $\frac{b-a}{a+b}>0$ since $a<b$, and $\cosh(x \log 2)>0$ since $\cosh$ is non-negative.
Therefore
$$f'(x)>0$$ 
as you're expecting.
