Mixing
Inequality of small two numbers with power
0
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
asked Nov 20 '18 at 19:29
MathRock
356
add a comment |
0
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
asked Nov 20 '18 at 19:29
MathRock
356
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
Nov 20 '18 at 19:32
I estimate some integration and on this, I face this problem.
– MathRock
Nov 20 '18 at 19:46
add a comment |
0
0
0
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
asked Nov 20 '18 at 19:29
MathRock
356
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
asked Nov 20 '18 at 19:29
MathRock
356
asked Nov 20 '18 at 19:29
MathRock
356
asked Nov 20 '18 at 19:29
MathRock
356
asked Nov 20 '18 at 19:29
MathRock
356
asked Nov 20 '18 at 19:29
MathRock
356
356
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
Nov 20 '18 at 19:32
I estimate some integration and on this, I face this problem.
– MathRock
Nov 20 '18 at 19:46
add a comment |
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
Nov 20 '18 at 19:32
I estimate some integration and on this, I face this problem.
– MathRock
Nov 20 '18 at 19:46
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
Nov 20 '18 at 19:32
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
Nov 20 '18 at 19:32
I estimate some integration and on this, I face this problem.
– MathRock
Nov 20 '18 at 19:46
I estimate some integration and on this, I face this problem.
– MathRock
Nov 20 '18 at 19:46
add a comment |
1 Answer
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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 $y=0?$
answered Nov 20 '18 at 19:36
Eric
2068
foy y=0 and y=1 it's not true as $x<1$
– MathRock
Nov 20 '18 at 19:38
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
Nov 20 '18 at 19:40
Can you tell more?
– MathRock
Nov 20 '18 at 19:41
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
Nov 20 '18 at 19:51
add a comment |
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1 Answer
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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 $y=0?$
answered Nov 20 '18 at 19:36
Eric
2068
foy y=0 and y=1 it's not true as $x<1$
– MathRock
Nov 20 '18 at 19:38
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
Nov 20 '18 at 19:40
Can you tell more?
– MathRock
Nov 20 '18 at 19:41
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
Nov 20 '18 at 19:51
add a comment |
0
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 $y=0?$
answered Nov 20 '18 at 19:36
Eric
2068
foy y=0 and y=1 it's not true as $x<1$
– MathRock
Nov 20 '18 at 19:38
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
Nov 20 '18 at 19:40
Can you tell more?
– MathRock
Nov 20 '18 at 19:41
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
Nov 20 '18 at 19:51
add a comment |
0
0
0
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 $y=0?$
answered Nov 20 '18 at 19:36
Eric
2068
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 $y=0?$
answered Nov 20 '18 at 19:36
Eric
2068
edited Nov 20 '18 at 19:40
answered Nov 20 '18 at 19:36
Eric
2068
answered Nov 20 '18 at 19:36
Eric
2068
answered Nov 20 '18 at 19:36
Eric
2068
2068
foy y=0 and y=1 it's not true as $x<1$
– MathRock
Nov 20 '18 at 19:38
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
Nov 20 '18 at 19:40
Can you tell more?
– MathRock
Nov 20 '18 at 19:41
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
Nov 20 '18 at 19:51
add a comment |
foy y=0 and y=1 it's not true as $x<1$
– MathRock
Nov 20 '18 at 19:38
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
Nov 20 '18 at 19:40
Can you tell more?
– MathRock
Nov 20 '18 at 19:41
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
Nov 20 '18 at 19:51
foy y=0 and y=1 it's not true as $x<1$
– MathRock
Nov 20 '18 at 19:38
foy y=0 and y=1 it's not true as $x<1$
– MathRock
Nov 20 '18 at 19:38
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
Nov 20 '18 at 19:40
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
Nov 20 '18 at 19:40
Can you tell more?
– MathRock
Nov 20 '18 at 19:41
Can you tell more?
– MathRock
Nov 20 '18 at 19:41
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
Nov 20 '18 at 19:51
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
Nov 20 '18 at 19:51
add a comment |
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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
Nov 20 '18 at 19:32
I estimate some integration and on this, I face this problem.
– MathRock
Nov 20 '18 at 19:46