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!

Using
$$
 \frac{a}{a+b} = \frac 12 \left( 1 - \frac{b-a}{a+b}\right) \, , \quad
\frac{b}{a+b} = \frac 12 \left( 1 + \frac{b-a}{a+b}\right)
$$
we have
$$
\begin{aligned}
\frac{a \cdot 2^x + b \cdot 2^{-x}}{a+b} &= \left(\frac{2^x + 2^{-x}}{2}\right) - \frac{b-a}{a+b} \left(\frac{2^x - 2^{-x}}{2}\right) \\
 &= \cosh(x \log 2) - \frac{b-a}{a+b} \sinh(x \log 2) \\
 &\le \cosh(x \log 2) \, .
\end{aligned}
$$

An alternative argument: Fix $x \ge 0$. For $0 \le t \le \frac 12$ we have
$$
 t \cdot e^x + (1-t) \cdot e^{-x} \le \max(e^{-x}, \frac{2^x + 2^{-x}}{2}) = \frac{2^x + 2^{-x}}{2}
$$
because the left-hand side is a linear function of $t$ which attains its maximum at a boundary point of the interval $[0, \frac 12]$. Setting $t = \frac{a}{a+b}$ gives the desired conclusion.
