Inequality of small two numbers with power











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Let $0<x,y<1$ are two real numbers and $ninmathbb N$. Is the following inequality true $$x^n-y^nleq x^{n+1}-y^{n+1}?$$



I split into two cases:



Case1: when $y<x$.



Case2:when $x<y.$



But in both the cases, I can't conclude anything.






real-analysis inequality real-numbers a.m.-g.m.-inequality







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  • It's enough to assume that $x geq y$, say. Have you considered some values of $x$ and $y$ to test if it's true? In what context did this question arise? What tools are available for use here?
    – JavaMan
    6 hours ago










  • I estimate some integration and on this, I face this problem.
    – MathRock
    6 hours ago















up vote
0
down vote

favorite












Let $0<x,y<1$ are two real numbers and $ninmathbb N$. Is the following inequality true $$x^n-y^nleq x^{n+1}-y^{n+1}?$$



I split into two cases:



Case1: when $y<x$.



Case2:when $x<y.$



But in both the cases, I can't conclude anything.






real-analysis inequality real-numbers a.m.-g.m.-inequality







share|cite|improve this question






















  • It's enough to assume that $x geq y$, say. Have you considered some values of $x$ and $y$ to test if it's true? In what context did this question arise? What tools are available for use here?
    – JavaMan
    6 hours ago










  • I estimate some integration and on this, I face this problem.
    – MathRock
    6 hours ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $0<x,y<1$ are two real numbers and $ninmathbb N$. Is the following inequality true $$x^n-y^nleq x^{n+1}-y^{n+1}?$$



I split into two cases:



Case1: when $y<x$.



Case2:when $x<y.$



But in both the cases, I can't conclude anything.






real-analysis inequality real-numbers a.m.-g.m.-inequality







share|cite|improve this question













Let $0<x,y<1$ are two real numbers and $ninmathbb N$. Is the following inequality true $$x^n-y^nleq x^{n+1}-y^{n+1}?$$



I split into two cases:



Case1: when $y<x$.



Case2:when $x<y.$



But in both the cases, I can't conclude anything.






real-analysis inequality real-numbers a.m.-g.m.-inequality




real-analysis inequality real-numbers a.m.-g.m.-inequality






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 6 hours ago









MathRock

306




306












  • It's enough to assume that $x geq y$, say. Have you considered some values of $x$ and $y$ to test if it's true? In what context did this question arise? What tools are available for use here?
    – JavaMan
    6 hours ago










  • I estimate some integration and on this, I face this problem.
    – MathRock
    6 hours ago


















  • It's enough to assume that $x geq y$, say. Have you considered some values of $x$ and $y$ to test if it's true? In what context did this question arise? What tools are available for use here?
    – JavaMan
    6 hours ago










  • I estimate some integration and on this, I face this problem.
    – MathRock
    6 hours ago
















It's enough to assume that $x geq y$, say. Have you considered some values of $x$ and $y$ to test if it's true? In what context did this question arise? What tools are available for use here?
– JavaMan
6 hours ago




It's enough to assume that $x geq y$, say. Have you considered some values of $x$ and $y$ to test if it's true? In what context did this question arise? What tools are available for use here?
– JavaMan
6 hours ago












I estimate some integration and on this, I face this problem.
– MathRock
6 hours ago




I estimate some integration and on this, I face this problem.
– MathRock
6 hours ago










1 Answer
1






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If its true for all $x,y$ with $0<x,y<1$, then it must also be true for all $x,y$ with $0 leq x, y leq 1$, since both $x^n-y^n$ and $x^{n+1}-y^{n+1}$ are continuous functions.



What happens if you let $y=0?$






share|cite|improve this answer























  • foy y=0 and y=1 it's not true as $x<1$
    – MathRock
    6 hours ago












  • Exactly, so if you prove the assertion that it also must hold for $x,y$ with $0 leq x,y leq 1$ then you are done.
    – Eric
    6 hours ago










  • Can you tell more?
    – MathRock
    6 hours ago










  • I claim that if $f(x,y)$ and $g(x,y)$ are continuous functions and you a sequences $(x_n,y_n)$ which converges to $(x,y)$ and $f(x_n,y_n) leq g(x_n, y_n)$ for all $n$ then you must have $f(x,y) leq g(x,y)$. If not then $f(x,y)-g(x,y)> epsilon >0$ for some $epsilon$. You should be able to use that and the continuity of $f$ and $g$ (hence the fact that $f(x_n, y_n) $ converges to $f(x,y)$ and likewise for $g$) to get a contradiction.
    – Eric
    6 hours ago











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1 Answer
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1 Answer
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active

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active

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up vote
0
down vote













If its true for all $x,y$ with $0<x,y<1$, then it must also be true for all $x,y$ with $0 leq x, y leq 1$, since both $x^n-y^n$ and $x^{n+1}-y^{n+1}$ are continuous functions.



What happens if you let $y=0?$






share|cite|improve this answer























  • foy y=0 and y=1 it's not true as $x<1$
    – MathRock
    6 hours ago












  • Exactly, so if you prove the assertion that it also must hold for $x,y$ with $0 leq x,y leq 1$ then you are done.
    – Eric
    6 hours ago










  • Can you tell more?
    – MathRock
    6 hours ago










  • I claim that if $f(x,y)$ and $g(x,y)$ are continuous functions and you a sequences $(x_n,y_n)$ which converges to $(x,y)$ and $f(x_n,y_n) leq g(x_n, y_n)$ for all $n$ then you must have $f(x,y) leq g(x,y)$. If not then $f(x,y)-g(x,y)> epsilon >0$ for some $epsilon$. You should be able to use that and the continuity of $f$ and $g$ (hence the fact that $f(x_n, y_n) $ converges to $f(x,y)$ and likewise for $g$) to get a contradiction.
    – Eric
    6 hours ago















up vote
0
down vote













If its true for all $x,y$ with $0<x,y<1$, then it must also be true for all $x,y$ with $0 leq x, y leq 1$, since both $x^n-y^n$ and $x^{n+1}-y^{n+1}$ are continuous functions.



What happens if you let $y=0?$






share|cite|improve this answer























  • foy y=0 and y=1 it's not true as $x<1$
    – MathRock
    6 hours ago












  • Exactly, so if you prove the assertion that it also must hold for $x,y$ with $0 leq x,y leq 1$ then you are done.
    – Eric
    6 hours ago










  • Can you tell more?
    – MathRock
    6 hours ago










  • I claim that if $f(x,y)$ and $g(x,y)$ are continuous functions and you a sequences $(x_n,y_n)$ which converges to $(x,y)$ and $f(x_n,y_n) leq g(x_n, y_n)$ for all $n$ then you must have $f(x,y) leq g(x,y)$. If not then $f(x,y)-g(x,y)> epsilon >0$ for some $epsilon$. You should be able to use that and the continuity of $f$ and $g$ (hence the fact that $f(x_n, y_n) $ converges to $f(x,y)$ and likewise for $g$) to get a contradiction.
    – Eric
    6 hours ago













up vote
0
down vote










up vote
0
down vote









If its true for all $x,y$ with $0<x,y<1$, then it must also be true for all $x,y$ with $0 leq x, y leq 1$, since both $x^n-y^n$ and $x^{n+1}-y^{n+1}$ are continuous functions.



What happens if you let $y=0?$






share|cite|improve this answer














If its true for all $x,y$ with $0<x,y<1$, then it must also be true for all $x,y$ with $0 leq x, y leq 1$, since both $x^n-y^n$ and $x^{n+1}-y^{n+1}$ are continuous functions.



What happens if you let $y=0?$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 6 hours ago

























answered 6 hours ago









Eric

915




915












  • foy y=0 and y=1 it's not true as $x<1$
    – MathRock
    6 hours ago












  • Exactly, so if you prove the assertion that it also must hold for $x,y$ with $0 leq x,y leq 1$ then you are done.
    – Eric
    6 hours ago










  • Can you tell more?
    – MathRock
    6 hours ago










  • I claim that if $f(x,y)$ and $g(x,y)$ are continuous functions and you a sequences $(x_n,y_n)$ which converges to $(x,y)$ and $f(x_n,y_n) leq g(x_n, y_n)$ for all $n$ then you must have $f(x,y) leq g(x,y)$. If not then $f(x,y)-g(x,y)> epsilon >0$ for some $epsilon$. You should be able to use that and the continuity of $f$ and $g$ (hence the fact that $f(x_n, y_n) $ converges to $f(x,y)$ and likewise for $g$) to get a contradiction.
    – Eric
    6 hours ago


















  • foy y=0 and y=1 it's not true as $x<1$
    – MathRock
    6 hours ago












  • Exactly, so if you prove the assertion that it also must hold for $x,y$ with $0 leq x,y leq 1$ then you are done.
    – Eric
    6 hours ago










  • Can you tell more?
    – MathRock
    6 hours ago










  • I claim that if $f(x,y)$ and $g(x,y)$ are continuous functions and you a sequences $(x_n,y_n)$ which converges to $(x,y)$ and $f(x_n,y_n) leq g(x_n, y_n)$ for all $n$ then you must have $f(x,y) leq g(x,y)$. If not then $f(x,y)-g(x,y)> epsilon >0$ for some $epsilon$. You should be able to use that and the continuity of $f$ and $g$ (hence the fact that $f(x_n, y_n) $ converges to $f(x,y)$ and likewise for $g$) to get a contradiction.
    – Eric
    6 hours ago
















foy y=0 and y=1 it's not true as $x<1$
– MathRock
6 hours ago






foy y=0 and y=1 it's not true as $x<1$
– MathRock
6 hours ago














Exactly, so if you prove the assertion that it also must hold for $x,y$ with $0 leq x,y leq 1$ then you are done.
– Eric
6 hours ago




Exactly, so if you prove the assertion that it also must hold for $x,y$ with $0 leq x,y leq 1$ then you are done.
– Eric
6 hours ago












Can you tell more?
– MathRock
6 hours ago




Can you tell more?
– MathRock
6 hours ago












I claim that if $f(x,y)$ and $g(x,y)$ are continuous functions and you a sequences $(x_n,y_n)$ which converges to $(x,y)$ and $f(x_n,y_n) leq g(x_n, y_n)$ for all $n$ then you must have $f(x,y) leq g(x,y)$. If not then $f(x,y)-g(x,y)> epsilon >0$ for some $epsilon$. You should be able to use that and the continuity of $f$ and $g$ (hence the fact that $f(x_n, y_n) $ converges to $f(x,y)$ and likewise for $g$) to get a contradiction.
– Eric
6 hours ago




I claim that if $f(x,y)$ and $g(x,y)$ are continuous functions and you a sequences $(x_n,y_n)$ which converges to $(x,y)$ and $f(x_n,y_n) leq g(x_n, y_n)$ for all $n$ then you must have $f(x,y) leq g(x,y)$. If not then $f(x,y)-g(x,y)> epsilon >0$ for some $epsilon$. You should be able to use that and the continuity of $f$ and $g$ (hence the fact that $f(x_n, y_n) $ converges to $f(x,y)$ and likewise for $g$) to get a contradiction.
– Eric
6 hours ago


















 

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