In Shiryaev's book problems in probability, an upper bound for binomial coefficients is shown in section 1.2.1: $$\binom{m+n}{n}\le(1+m/n)^n(1+n/m)^m.$$ It seems that this bound is sharper than the well known result that $\binom{n}{k}\le (ne/k)^k$ because $(1+n/m)^m = (1+m/n)^{(m/n)n} < e^n$.
Could anyone tell me how to derive this tighter bound?
If you multiply the inequality by $n^nm^m$, the RHS can be expanded with the binomial theorem, and the LHS will be but one of the terms of the resulting sum.
If you want a tighter bound.
Using my answer here
$$\binom{m+n}{n}\le \frac{1}{\sqrt{2 \pi }}\sqrt{\frac{1}{m}+\frac{1}{n}}\,\frac{(m+n)^{m+n} }{m^m\, n^n }$$ Comparing
$$\frac{\frac{1}{\sqrt{2 \pi }}\sqrt{\frac{1}{m}+\frac{1}{n}}\,\frac{(m+n)^{m+n} }{m^m\, n^n } } {\left(1+\frac{m}{n}\right)^n \left(1+\frac{n}{m}\right)^m }=\sqrt{\frac{\frac{1}{m}+\frac{1}{n}}{2 \pi }} \ll1$$
