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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.

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

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    $\begingroup$ Bernoulli's inequality also gives $(1+m)\le\left(1+\frac mn\right)^n$, and I was writing that up when your post appeared. I found another approach, though not as neat. (+1) $\endgroup$ – robjohn Jul 29 '15 at 0:21
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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.

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    $\begingroup$ There are $m+n$ terms and not $m\cdot n$ terms on the top. $\endgroup$ – DeepSea Jul 28 '15 at 21:59
  • $\begingroup$ @KfSsOc I think in post all is okay. $\endgroup$ – Dmytro Chasovskyi Jul 28 '15 at 22:08
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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 $m,n\ge1$.

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