Struggling with an inequality: $ \frac{1}{\sqrt[n]{1+m}} + \frac{1}{\sqrt[m]{1+n}} \ge 1 $ [duplicate]

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.

Solution 1:

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: $$ \frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}}\geq \frac{1}{1+\frac{m}{n}}+\frac{1}{1+\frac{n}{m}}=1.$$

Solution 2:

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.