Examining an inequality involving exponential functions and hyperbolic cosine

Question

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!

Answer 1

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.

Answer 2

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.

Answer 3

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

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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.