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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?

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    $\begingroup$ $$\sqrt[x+3]x $$ is a decreasing function $\endgroup$ – lab bhattacharjee Dec 27 '18 at 7:47
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    $\begingroup$ @labbhattacharjee gives a good hint. Just analyse $f(x) = \sqrt[x+3]{x}$. $\endgroup$ – Matti P. Dec 27 '18 at 7:53
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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Dec 27 '18 at 8:08
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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Dec 27 '18 at 8:08
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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Dec 27 '18 at 8:24
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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 $e<2.72\implies e^2<7.3984<7.4\implies$ $\implies e^4<7.4^2=54.76<55\implies$ $e^8<55^2=3025<3125=5^5.$

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    $\begingroup$ We can also obtain $g(5)<0$ by knowing that $\ln 10>2.3$ and $\ln 2<0.7,$ as $g(5)=8-5(\ln 10-\ln 2)$. $\endgroup$ – DanielWainfleet Dec 27 '18 at 8:44
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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 $(1+\frac 1 n)^{n+3} <n$. 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 $1+\frac 3 n \leq \frac 8 5 < \ln 5 \leq \ln \, n$ . [$e^{1.6} =4.965302...$].

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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 $\left(1+\frac{1}{n}\right)^n<e,$ 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!

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