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!
 A: 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.
A: 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.
A: 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.
