When does this inequality hold: $h^2/\beta^2 \le \beta \exp(h/ \beta)$?

Question

If $h<0$ and $\beta >0$. When we can have the next inequality $$\frac{h^2}{\beta^2} \le \beta \cdot\exp(h/ \beta)$$

Answer 1

You won't be able to solve $$ \frac{h^2}{\beta^2} \le \beta \cdot\exp(\frac h \beta) $$ for $h<0,\beta>0$ in a general algebraic way in the sense that you get anything like $$ f(h)\le g(\beta) $$ with $f,g$ are some suitable functions. Of course you could try to use some numerical methods, for example Wolframalpha produces a nice plot.

However, to get an intuition it might be the best to go for three cases $$ |h|>|\beta|,\,|h|<|\beta|,\,|h|=|\beta| $$ You'll find for each case solutions but for the third case, when we have $|h|=|\beta|,h<0,\beta>0$, we can indeed construct a solution since $$ |h|=|\beta|\implies\frac {h^2}{\beta^2}=1,\frac {h}{\beta}=-1 $$ which gives us $$ \frac{h^2}{\beta^2} \le \beta \cdot\exp(\frac h \beta)\iff1\leq\beta\frac1 e\iff e\leq\beta $$ but that's about it.