Prove that $\sqrt[8]5 > \sqrt[9]6 > \sqrt[10]7 > \cdots$

Prove that $$\sqrt[8]5 > \sqrt[9]6 > \sqrt[10]7 > \cdots$$

My friend came up with this and gave this to me as a challenge and I'm totally stuck.

I have tried proving this by induction $$\root{n+3}\of{n} > \root{n+4} \of {n+1}$$ for all integers $$n \geq 5$$ with no luck. I don't even know how to prove the base case without a calculator. Also, it turns out that this is not true for $$n \leq 4$$. Why would this inequality only true from $$5$$ onwards?

• $$\sqrt[x+3]x$$ is a decreasing function – lab bhattacharjee Dec 27 '18 at 7:47
• @labbhattacharjee gives a good hint. Just analyse $f(x) = \sqrt[x+3]{x}$. – Matti P. Dec 27 '18 at 7:53
• See this thread for cues. As you tagged this calculus presumably using derivatives is allowed, so you can look at Yves Daoust's answer in that thread in particular. – Jyrki Lahtonen Dec 27 '18 at 8:08
• Anyway $$Dx^{1/(x+3)}=-\frac{x^{\frac{1}{x+3}-1} (-x+x \log (x)-3)}{(x+3)^2}.$$ Looking at that form it is clear which factor determines the sign. – Jyrki Lahtonen Dec 27 '18 at 8:08
• On second thought, you do need careful estimates to prove that the derivative above is negative already at $x=5$. After all, $5\ln 5-8$ is rather close to zero. Anyway, the dervative will take care of the infinite tail of the inequalities. – Jyrki Lahtonen Dec 27 '18 at 8:24

Let $$f(x)=\frac {1}{x+3}\ln x .$$ Then $$f'(x)=\frac {x+3-x\ln x}{x(x+3)}.$$

Let $$g(x)=x+3-x\ln x.$$ Then $$g'(x)=-\ln x.$$ Now $$g(x)$$ is strictly decreasing for $$x\geq 5$$ because $$g'(x)<0$$ for $$x\geq 5.$$ By calculation $$g(5)<0.$$ So $$g(x)\leq g(5)<0$$ for $$x\geq 5.$$

Therefore $$f'(x)=\frac {g(x)}{x(x+3)}<0$$ for $$x\geq 5,$$ so $$f(x)$$ is strictly decreasing for $$x\geq 5.$$ So $$e^{f(x)}=x^{1/(x+3)}$$ is strictly decreasing for $$x\ge 5.$$

Remark: $$g(x)$$ is strictly decreasing for $$x\geq 1$$ but for small $$x>1$$ we have $$g(x)>0$$. E.g. $$g(4)>0$$. And $$5$$ is the least $$n\in \Bbb N$$ such that $$g(n)<0$$, i.e. such that $$f'(n)<0.$$

Remark. $$g(5)<0\iff e^8<5^5.$$ We have $$e<2.72\implies e^2<7.3984<7.4\implies$$ $$\implies e^4<7.4^2=54.76<55\implies$$ $$e^8<55^2=3025<3125=5^5.$$

• We can also obtain $g(5)<0$ by knowing that $\ln 10>2.3$ and $\ln 2<0.7,$ as $g(5)=8-5(\ln 10-\ln 2)$. – DanielWainfleet Dec 27 '18 at 8:44

What yo need is $$n^{\frac 1 {n+3}} >(n+1)^{\frac 1 {n+4}}$$ which is same as $$n^{n+4} >(n+1)^{n+3}$$ or $$(1+\frac 1 n)^{n+3} . In other words you need $$(n+3)\ln \, (1+\frac 1 n) <\ln\, n$$Since $$\ln \, (1+\frac 1 n) <\frac 1 n$$ it is enough to show that $$\frac {n+3} n <\ln\, n$$ or $$1+\frac 3 n <\ln\, n$$. Since we have $$n \geq 5$$ we have $$1+\frac 3 n \leq \frac 8 5 < \ln 5 \leq \ln \, n$$ . [$$e^{1.6} =4.965302...$$].

For $$n\geq5$$ we need to prove that $$n^{\frac{1}{n+3}}>(n+1)^{\frac{1}{n+4}}$$ or $$n^{n+4}>(n+1)^{n+3}$$ or $$\frac{n^4}{(n+1)^3}>\left(1+\frac{1}{n}\right)^n$$ and since $$\left(1+\frac{1}{n}\right)^n it's enough to prove that $$n^4-e(n+1)^3>0.$$ But the polynomial $$n^4-e(n+1)^3$$ has one changing of coefficients signs.

Thus, by the Descartes's rule this polynomial has unique positive root.

Id est, it remains to check that $$\frac{5^4}{6^3}>e$$ and since $$e=2.718...<2.75,$$ it's enough to prove that $$\frac{625}{216}>2.75$$ or $$625>594.$$ Done!