# Question about a pure algebraic proof

it's related to this Proving $\ln \cosh x\leq \frac{x^2}{2}$ for $x\in\mathbb{R}$

The case $x\geq 2$ is easy so continue with :

The case $0\leq x \leq 2$

Purely "algebraic" proof of Young's Inequality

If we put $a=e^{2x}$$\quad$$b=e^{-x}$$\quad$$p=\frac{x}{4}$ we get: $$\frac{4e^{0.5x^2}}{x}+\frac{e^{\frac{-x^2}{x-4}}}{\frac{x}{x-4}}\geq e^x$$ Or : $$e^{0.5x^2}\geq (e^x-e^{\frac{-x^2}{x-4}}\frac{x-4}{x})\frac{x}{4}$$ Or : $$e^{0.5x^2}\geq e^x(\frac{x}{4}) -(e^{\frac{-x^2}{x-4}})\frac{x-4}{x}\frac{x}{4}$$

Futhermore it's easy to remark that we have :