Prove that $\frac{1}{^n\sqrt{1+m}}+\frac{1}{^m\sqrt{1+n}}\geq1$

hint: $\sqrt[n]{1+m} =\sqrt[n]{1\cdot 1\cdot 1\cdots 1\cdot (1+m)} \leq \dfrac{n-1+1+m}{n}=\dfrac{m+n}{n}$

In fact,your inequality for any $m,n\in R(>1)$ is hold.

because you can use Bernoulli inequality $$(1+x)^a\le 1+ax,0<a<1\Longrightarrow \sqrt[n]{1+m}=(1+m)^{\frac{1}{n}}<1+\frac{m}{n}$$ then $$\dfrac{1}{\sqrt[n]{m+1}}>\dfrac{1}{1+\frac{m}{n}}=\dfrac{n}{m+n}$$