Struggling with an inequality: $ \frac{1}{\sqrt[n]{1+m}} + \frac{1}{\sqrt[m]{1+n}} \ge 1 $ Prove that for every natural numbers, $m$ and $n$, this inequality holds:
$$
\frac{1}{\sqrt[n]{1+m}} +  \frac{1}{\sqrt[m]{1+n}} \ge 1
$$
I tried to use Bernoulli's inequality, but I can't figure it out.

The AM-GM inequality says that for $x,y\ge0$ and $0\le a\le1$,
$$
ax+(1-a)y\ge x^ay^{1-a}\tag{1}
$$
substituting $x\mapsto x/a$ and $y\mapsto y/(1-a)$ yields
$$
x+y\ge\frac{x^ay^{1-a}}{a^a(1-a)^{1-a}}\tag{2}
$$
Therefore, with $x=(m+1)^{-1/n}$, $y=(n+1)^{-1/m}$, and $a=\frac n{m+n}$, we get
$$
\begin{align}
(m+1)^{-1/n}+(n+1)^{-1/m}
&\ge\frac{[(m+1)(n+1)]^{-\frac1{m+n}}}{\left(\frac m{m+n}\right)^{\frac m{m+n}}\left(\frac n{m+n}\right)^{\frac n{m+n}}}\\[6pt]
&=\frac{m+n}{\left[(m+1)m^m(n+1)n^n\right]^{\frac1{m+n}}}\\[12pt]
&\ge1\tag{3}
\end{align}
$$

The last step in $(3)$ is equivalent to
$$
(m+n)^{m+n}\ge(m+1)(n+1)m^mn^n\tag{4}
$$
Note that $(4)$ is true when $m=n=1$. Suppose that $(4)$ is true for some $m,n$, then as $m$ or $n$ is increased by $1$, the left side of $(4)$ increases by a factor of
$$
\frac{(m+n+1)^{m+n+1}}{(m+n)^{m+n}}
=\color{#C00000}{(m+n+1)}\color{#00A000}{\left(1+\frac1{m+n}\right)^{m+n}}\tag{5}
$$
and if $n$ is increased by $1$ (and similarly for $m$), the right side of $(4)$ increases by a factor of
$$
\frac{(n+2)(n+1)^{n+1}}{(n+1)n^n}
=\color{#C00000}{(n+2)}\color{#00A000}{\left(1+\frac1n\right)^n}\tag{6}
$$
Since $m,n\ge1$, we have that $\color{#C00000}{m+n+1\ge n+2}$, and Bernoulli's Inequality ensures that
$\color{#00A000}{\left(1+\frac1{m+n}\right)^{m+n}\ge\left(1+\frac1n\right)^n}$.
Comparing $(5)$ and $(6)$ ensures that inequality $(4)$ remains valid for all $m,n\ge1$.
