How to show that this inequality about exponential is false?

Question

I want to show that $$e^h(e^h-1-h)h^2\geq 2(e^h-1-he^h)^2$$ is false for sufficiently small $h>0$.

I tried Taylor expansion, but I wasn't sure how to proceed. Any hint is greatly appreciated.

Answer 1

Using the classical Taylor expansion of $e^h$, we have $$e^h \left(e^h-1-h\right) h^2=\frac{h^4}{2}+\frac{2 h^5}{3}+\frac{11 h^6}{24}+\frac{13 h^7}{60}+O\left(h^8\right)$$ $$2 \left(e^h-1-h e^h \right)^2=\frac{h^4}{2}+\frac{2 h^5}{3}+\frac{17 h^6}{36}+\frac{7 h^7}{30}+O\left(h^8\right)$$ $$\Delta=e^h \left(e^h-1-h\right) h^2-2 \left(e^h-1-h e^h \right)^2=-\frac{h^6}{72}-\frac{h^7}{60}+O\left(h^8\right)$$

In fact, pushing the expansion up to $O\left(h^{101}\right)$, only negative terms appear in the expansion of $\Delta$.

Combining the coefficients $$\Delta =-\sum_{n=6}^\infty \frac{2^n (n-8)+4 (n^2-n+4)}{4 n (n-2)!}h^n$$

Further computations show that the maximum value of $\Delta$ is $0$ (at $h=0$).

Then the inequality is always false for any $h \neq 0$.