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.
 A: If we change a little this inequality we will have:
$$ \sqrt[m \cdot n]{(1 + n )^n} + \sqrt[m \cdot n]{(1 + m)^m} \geq \sqrt[m \cdot n]{(1+n)^n \cdot (1+m)^m}$$
From the first part we will have:
$$ \sqrt[m \cdot n]{(1 + n )^n} + \sqrt[m \cdot n]{(1 + m)^m} \geq 1 + \frac{n}{m} + 1 + \frac{m}{n} = \frac{(m+n)^2}{m \cdot n}$$
From tail we have inequality( using this fact Inequality of arithmetic and geometric means):
$$ \sqrt[m \cdot n]{(1+n)^n \cdot (1+m)^m} \leq \frac{ n \cdot (1+n) + m \cdot ( 1 + m)}{m \cdot n} = \frac{(n+m)^2 - 2\cdot m \cdot n + m + n }{m \cdot n}$$
As a result we have:
$ \frac{(n+m)^2 }{m \cdot n} \geq \frac{(n+m)^2 - 2\cdot m \cdot n + m + n }{m \cdot n}$
or
$$ 2\cdot m \cdot n \geq m + n $$ that is true, because $m$ and $n$ are natural.  
A: 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 $m,n\ge1$.
A: The inequality is fulfilled if $mn=0$, so we may assume $m,n\neq 0$ without loss of generality. 
By the AM-GM inequality:
$$\sqrt[n]{1+m}\leq \frac{1+\ldots+1+(m+1)}{n} = 1+\frac{m}{n},$$
hence:
$$ \frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}}\geq \frac{1}{1+\frac{m}{n}}+\frac{1}{1+\frac{n}{m}}=1.$$
