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This question already has an answer here:

I wish to prove the following inequality:

$$\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}}\ge 1$$

I tried to use different ideas, I tried using squeeze theorem or assuming $m\gt n$ and working my way from there, but I did not manage to prove this inequality.

Any suggestions?

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marked as duplicate by gimusi calculus Oct 30 '18 at 17:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Prove $(1+m)^(1/n)\leq 1+m/n$ first. $\endgroup$ – user10354138 Oct 30 '18 at 17:29
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Well, $(1+m)^{1/n} \le (1+m/n)$, and $(1+n)^{1/m} \le (1+n/m)$. [Do you see why?]

From this the inequality follows:

$$\frac{1}{(1+m)^{1/n}} + \frac{1}{(1+n)^{1/m}} \ge \frac{1}{1+\frac{m}{n}} + \frac{1}{1+\frac{n}{m}} = \frac{n}{n+m} + \frac{m}{n+m} = 1$$

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  • $\begingroup$ A way to see the inequality $(1+m)^{1/n} \le 1+m/n$ for $n > 1$ is this. Let $f(x) = (1+m)^x$ and let $g(x) = 1+xm$. Then $f(0)=g(0)$ and $f(1)=g(1)$. However, $f$ is convex in $x$, so for each $x \in [0,1]$ it follows that $g(x) \le f(x)$. Take Then take $x=\frac{1}{n}$. $\endgroup$ – Mike Oct 30 '18 at 18:18
  • $\begingroup$ We do need to make some restrictions on $m$ and $n$ (I assumed $m,n \geq 1$). If $m=n=\frac{1}{3}$ then the above inequality does not hold. $\endgroup$ – Mike Oct 30 '18 at 18:20
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For any $x>-1$ and $s\in[0,1]$, we have $(1+x)^s\leq1+sx$ by Bernoulli's Inequality (with equality cases $s\in\{0,1\}$ and $x=0$). Ergo, if $m$ and $n$ are any real numbers greater than or equal to $1$ (not just positive integers as the notation seems to suggest), then $$\sqrt[m]{1+n}=(1+n)^{\frac1m}\leq 1+\frac{n}{m}$$ and $$\sqrt[n]{1+m}=(1+m)^{\frac1n}\leq 1+\frac{m}{n}\,.$$ The rest is just as the answer by Mike. However, note that the unique equality case is when $(m,n)=(1,1)$.

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  • $\begingroup$ That is for $s<0$ and $s>1$. The inequality is flipped for $0\leq s\leq 1$. $\endgroup$ – Batominovski Oct 30 '18 at 17:41
  • $\begingroup$ Right, so then you're using a variant of the well known Bernoulli inequality and perhaps it'd be a good idea to mention it $\endgroup$ – DonAntonio Oct 30 '18 at 17:42
  • $\begingroup$ Well, it is on the Wiki page (stated in the introduction, to be exact). But ok, I will put the link in there. $\endgroup$ – Batominovski Oct 30 '18 at 17:43
  • $\begingroup$ Yes implicit in the assumption in my answer is that $m,n > 1$. Now if $m=n=\frac{1}{3}$, then the sum becomes $2/(4/3)^3 < 1$. So we do need to make some assumptions on $m$ and $n$. Nice answer above @Batominovski ! $\endgroup$ – Mike Oct 30 '18 at 18:09
  • $\begingroup$ A way to see the inequality $(1+m)^{1/n} \le 1+m/n$ is this. Let $f(x) = (1+m)^x$ and let $g(x) = 1+xm$. Then $f(0)=g(0)$ and $f(1)=g(1)$. However, $f$ is convex in $x$, so for each $x \in [0,1]$ it follows that $g(x) \le f(x)$. Take Then take $x=\frac{1}{n}$. $\endgroup$ – Mike Oct 30 '18 at 18:14

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