Prove the inequality $\frac{e^x+e^{-x}}{2} \leq e^{x^2/2}$ for all real numbers $x$.

Your computation is flawed. The expansion of $\cosh{x}$ is $\sum_{j \geq 0}{\frac{x^{2j}}{(2j)!}}$, but the expansion of $e^{x^2/2}$ is $\sum_{j \geq 0}{\frac{x^{2j}}{2^j \cdot j!}}$.

So you just need to prove that $j!2^j < (2j)!$ for each $j$.


Let $f(x)=\frac{x^2}{2}-\ln\frac{e^x+e^{-x}}{2}.$

Since $f$ is an even function, it's enough to prove that $f(x)\geq0$ for $x\geq0.$

We see that $$f'(x)=x-\frac{e^x-e^{-x}}{e^x+e^{-x}}$$ and $$f''(x)=1-\frac{(e^x+e^{-x})^2-(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=\frac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2}\geq0.$$ Id est, $$f'(x)\geq f'(0)=0,$$ $$f(x)\geq f(0)=0$$ and we are done!

Tags:

Inequality

Real Analysis

Taylor Expansion

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