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












2















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?






inequality radicals







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  • 6




    $$sqrt[x+3]x $$ is a decreasing function
    – lab bhattacharjee
    Dec 27 '18 at 7:47






  • 1




    @labbhattacharjee gives a good hint. Just analyse $f(x) = sqrt[x+3]{x}$.
    – Matti P.
    Dec 27 '18 at 7:53






  • 1




    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






  • 1




    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








  • 2




    On second thought, you do need careful estimates to prove that the derivative above is negative already at $x=5$. After all, $5ln 5-8$ 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


















2















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?






inequality radicals







share|cite|improve this question


















  • 6




    $$sqrt[x+3]x $$ is a decreasing function
    – lab bhattacharjee
    Dec 27 '18 at 7:47






  • 1




    @labbhattacharjee gives a good hint. Just analyse $f(x) = sqrt[x+3]{x}$.
    – Matti P.
    Dec 27 '18 at 7:53






  • 1




    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






  • 1




    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








  • 2




    On second thought, you do need careful estimates to prove that the derivative above is negative already at $x=5$. After all, $5ln 5-8$ 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
















2












2








2


2






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?






inequality radicals







share|cite|improve this question














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?






inequality radicals




inequality radicals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 27 '18 at 7:43









Mint

5131416




5131416








  • 6




    $$sqrt[x+3]x $$ is a decreasing function
    – lab bhattacharjee
    Dec 27 '18 at 7:47






  • 1




    @labbhattacharjee gives a good hint. Just analyse $f(x) = sqrt[x+3]{x}$.
    – Matti P.
    Dec 27 '18 at 7:53






  • 1




    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






  • 1




    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








  • 2




    On second thought, you do need careful estimates to prove that the derivative above is negative already at $x=5$. After all, $5ln 5-8$ 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
















  • 6




    $$sqrt[x+3]x $$ is a decreasing function
    – lab bhattacharjee
    Dec 27 '18 at 7:47






  • 1




    @labbhattacharjee gives a good hint. Just analyse $f(x) = sqrt[x+3]{x}$.
    – Matti P.
    Dec 27 '18 at 7:53






  • 1




    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






  • 1




    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








  • 2




    On second thought, you do need careful estimates to prove that the derivative above is negative already at $x=5$. After all, $5ln 5-8$ 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










6




6




$$sqrt[x+3]x $$ is a decreasing function
– lab bhattacharjee
Dec 27 '18 at 7:47




$$sqrt[x+3]x $$ is a decreasing function
– lab bhattacharjee
Dec 27 '18 at 7:47




1




1




@labbhattacharjee gives a good hint. Just analyse $f(x) = sqrt[x+3]{x}$.
– Matti P.
Dec 27 '18 at 7:53




@labbhattacharjee gives a good hint. Just analyse $f(x) = sqrt[x+3]{x}$.
– Matti P.
Dec 27 '18 at 7:53




1




1




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




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




1




1




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






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






2




2




On second thought, you do need careful estimates to prove that the derivative above is negative already at $x=5$. After all, $5ln 5-8$ 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






On second thought, you do need careful estimates to prove that the derivative above is negative already at $x=5$. After all, $5ln 5-8$ 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












3 Answers
3






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oldest

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2














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



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



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



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



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






share|cite|improve this answer



















  • 1




    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



















2














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 $1+frac 3 n leq frac 8 5 < ln 5 leq ln , n$ . [$e^{1.6} =4.965302...$].






share|cite|improve this answer





























    1














    For $ngeq5$ 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!






    share|cite|improve this answer























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      3 Answers
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      active

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      3 Answers
      3






      active

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      active

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      active

      oldest

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      2














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



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



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



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



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






      share|cite|improve this answer



















      • 1




        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
















      2














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



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



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



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



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






      share|cite|improve this answer



















      • 1




        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














      2












      2








      2






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



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



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



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



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






      share|cite|improve this answer














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



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



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



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



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







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 27 '18 at 8:38

























      answered Dec 27 '18 at 8:15









      DanielWainfleet

      34.1k31647




      34.1k31647








      • 1




        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














      • 1




        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








      1




      1




      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




      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











      2














      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 $1+frac 3 n leq frac 8 5 < ln 5 leq ln , n$ . [$e^{1.6} =4.965302...$].






      share|cite|improve this answer


























        2














        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 $1+frac 3 n leq frac 8 5 < ln 5 leq ln , n$ . [$e^{1.6} =4.965302...$].






        share|cite|improve this answer
























          2












          2








          2






          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 $1+frac 3 n leq frac 8 5 < ln 5 leq ln , n$ . [$e^{1.6} =4.965302...$].






          share|cite|improve this answer












          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 $1+frac 3 n leq frac 8 5 < ln 5 leq ln , n$ . [$e^{1.6} =4.965302...$].







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 27 '18 at 7:53









          Kavi Rama Murthy

          50.5k31854




          50.5k31854























              1














              For $ngeq5$ 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!






              share|cite|improve this answer




























                1














                For $ngeq5$ 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!






                share|cite|improve this answer


























                  1












                  1








                  1






                  For $ngeq5$ 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!






                  share|cite|improve this answer














                  For $ngeq5$ 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!







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 27 '18 at 10:12

























                  answered Dec 27 '18 at 10:04









                  Michael Rozenberg

                  96.5k1589187




                  96.5k1589187






























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