I was working on a problem and reduced it to showing the following inequality:
$$2x e^{x^2/6} \ge e^x - e^{-x} \text{ for $x \ge 0$}$$
I tried expanding everything in Taylor series to no avail. I also tried defining the function $f(x):= 2xe^{x^2/6} - e^x + e^{-x}$, showing $f(0) = 0$ and trying to show $f'(x) \ge 0$ for $x \ge 0$, but I couldn't show the last part.
Is there something easy I'm missing?
The Taylor series approach works if, for every $k\geqslant0$, $a_k\geqslant b_k$, with $$ a_k=\frac1{k!6^k},\qquad b_k=\frac1{(2k+1)!}. $$ Now, $a_0=b_0=1$ and, for every $k\geqslant1$, $$ \frac{a_k}{b_k}=\frac{2k(2k+1)}{6k}\cdot\frac{a_{k-1}}{b_{k-1}}=\frac{2k+1}3\cdot\frac{a_{k-1}}{b_{k-1}}\geqslant\frac{a_{k-1}}{b_{k-1}}, $$ hence the result holds.