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!

Big wall of algebra incoming:
\begin{align*}
\frac{a\cdot2^{x}+b\cdot2^{-x}}{a+b}\le\cosh\left(x\ln2\right)&\implies\frac{a\cdot2^{x}+\frac{b}{2^{x}}}{a+b}\le\frac{2^{x}+2^{-x}}{2}\\
&\implies\frac{a\cdot2^{x}+\frac{b}{2^{x}}}{a+b}\le\frac{2^{x}+\frac{1}{2^{x}}}{2}\\
&\implies\frac{a\cdot2^{2x}+b}{2^{x}\left(a+b\right)}\le\frac{2^{2x}+1}{2\cdot 2^{x}}\\
&\implies\frac{2^{2x}+b}{a+b}\le\frac{2^{2x}+1}{2}\\
&\implies\frac{a\cdot4^{x}+b}{a+b}\le\frac{4^{x}+1}{2}\\
&\implies\frac{a\cdot4^{x}+b}{a+b}-\frac{4^{x}+1}{2}\le0\\
&\implies\frac{2\left(a\cdot4^{x}+b\right)-\left(a+b\right)\left(4^{x}+1\right)}{2\left(a+b\right)}\le0\\
&\implies\frac{2a\cdot4^{x}+2b-a\cdot4^{x}-b\cdot4^{x}-a-b}{2\left(a+b\right)}\le0\\
&\implies\frac{4^{x}\left(2a-a-b\right)+b-a}{2\left(a+b\right)}\le0\\
&\implies\frac{4^{x}\left(a-b\right)+b-a}{2\left(a+b\right)}\le0\\
&\implies\frac{4^{x}\left(a-b\right)-\left(a-b\right)}{2\left(a+b\right)}\le0\\
&\implies\frac{\left(a-b\right)\left(4^{x}-1\right)}{2\left(a+b\right)}\le0\\
&\implies\left(a-b\right)\left(4^{x}-1\right)\le0\\
&\implies4^{x}-1\ge0\text{ since }a-b<0\\
&\implies4^{x}\ge1,
\end{align*}
which is true for all $x\ge0$.
If someone can find a pretty way to format this, then feel free to edit it.
