# proving the inequality $\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}}\ge 1$ [duplicate]

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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• Prove $(1+m)^(1/n)\leq 1+m/n$ first. – user10354138 Oct 30 '18 at 17:29

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:

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

• 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 $x=\frac{1}{n}$. – Mike Oct 30 '18 at 18:18
• 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. – Mike Oct 30 '18 at 18:20

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 $$(m,n)=(1,1)$$.

• That is for $s<0$ and $s>1$. The inequality is flipped for $0\leq s\leq 1$. – Batominovski Oct 30 '18 at 17:41
• 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 – DonAntonio Oct 30 '18 at 17:42
• Well, it is on the Wiki page (stated in the introduction, to be exact). But ok, I will put the link in there. – Batominovski Oct 30 '18 at 17:43
• 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 ! – Mike Oct 30 '18 at 18:09
• 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 $x=\frac{1}{n}$. – Mike Oct 30 '18 at 18:14