cosh x inequality

Solution 1:

The infinite product representation of the hyperbolic cosine function gives $$\cosh(x)=\prod_{k=1}^\infty\left(1+{4x^2\over \pi^2(2k-1)^2}\right) \leq \exp\left(\sum_{k=1}^\infty {4x^2\over \pi^2(2k-1)^2}\right) = \exp(x^2/2).$$

Solution 2:

Hint: the wanted inequality is equivalent to $\ln(\cosh x) \leq \ln(e^{x^2/2}) $ which is in turn equivalent to $$\ln(\cosh x) \leq {x^2/2}.$$ Now define this function $f(x)=\ln(\cosh x) - {x^2/2}$ and $f'(x)=\tanh(x)-x$ and find maximum of $f$.

Solution 3:

Noticing that $\frac{d}{dx}\tanh x = 1 -\tanh^2x$, note that:

• $1-\tanh^2 u\le1\implies\int_0^t(1-\tanh^2 u)du\le\int_0^tdu\implies\tanh t \le t$
• $\int_0^x \tanh t dt\le\int_0^xtdt\implies\log\cosh x \le \frac{1}{2}x^2\implies\cosh x\le e^{\frac{1}{2}x^2}$