Prove inequality $y^(n-1)+x^nleq x^(n-1)+y^n$

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Is there a short way to show that
$$y^n-1+x^nleq x^n-1+y^n$$
for $ninmathbb N$ and $1<x<y$?
algebra-precalculus inequality
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up vote
-1
down vote
favorite
Is there a short way to show that
$$y^n-1+x^nleq x^n-1+y^n$$
for $ninmathbb N$ and $1<x<y$?
algebra-precalculus inequality
Anything wrong with the question (because of the downvote)?
– EuklidAlexandria
Sep 10 at 16:06
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up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Is there a short way to show that
$$y^n-1+x^nleq x^n-1+y^n$$
for $ninmathbb N$ and $1<x<y$?
algebra-precalculus inequality
Is there a short way to show that
$$y^n-1+x^nleq x^n-1+y^n$$
for $ninmathbb N$ and $1<x<y$?
algebra-precalculus inequality
algebra-precalculus inequality
asked Sep 10 at 8:35
EuklidAlexandria
3019
3019
Anything wrong with the question (because of the downvote)?
– EuklidAlexandria
Sep 10 at 16:06
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Anything wrong with the question (because of the downvote)?
– EuklidAlexandria
Sep 10 at 16:06
Anything wrong with the question (because of the downvote)?
– EuklidAlexandria
Sep 10 at 16:06
Anything wrong with the question (because of the downvote)?
– EuklidAlexandria
Sep 10 at 16:06
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
$f(x)=x^n-x^n-1$ is an increasing function on $(1,infty)$ because $f'(x)=nx^n-1-(n-1)x^n-2 >0$ ( since $x >1 >1-frac 1 n$). The required inequality simply says $x<y$ implies $f(x) <f(y)$.
add a comment |Â
up vote
3
down vote
$$1<x<yimplies 0<x-1<y-1.$$
Multiply the left inequalities $n-1$ times and the right ones once to get
$$x^n-1(x-1)<y^n-1(y-1).$$
Equality is not possible.
add a comment |Â
up vote
2
down vote
Need to show:
$x^n -x^n-1 lt y^n-y^n-1$;
$x^n-1(x-1) lt y^n-1(y-1)$.
Since $1<x<y$ :
$(y-1)>(x-1)>0$, and
$y^n-1 > x^n-1>0$(why?).
add a comment |Â
up vote
2
down vote
The inequality is equivalent to
$$
x^n-1(x-1)leq y^n-1(y-1)
$$
This inequality again follows directly from the condition $1<x<y$ since all terms are positive (this is where you need the condition $x,y>1$).
The method is good, but you are on the wrong side, $1<x<y$.
– Yves Daoust
Sep 10 at 8:50
@YvesDaoust It should be correct by now.
– Janik
Sep 10 at 8:55
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$f(x)=x^n-x^n-1$ is an increasing function on $(1,infty)$ because $f'(x)=nx^n-1-(n-1)x^n-2 >0$ ( since $x >1 >1-frac 1 n$). The required inequality simply says $x<y$ implies $f(x) <f(y)$.
add a comment |Â
up vote
3
down vote
accepted
$f(x)=x^n-x^n-1$ is an increasing function on $(1,infty)$ because $f'(x)=nx^n-1-(n-1)x^n-2 >0$ ( since $x >1 >1-frac 1 n$). The required inequality simply says $x<y$ implies $f(x) <f(y)$.
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$f(x)=x^n-x^n-1$ is an increasing function on $(1,infty)$ because $f'(x)=nx^n-1-(n-1)x^n-2 >0$ ( since $x >1 >1-frac 1 n$). The required inequality simply says $x<y$ implies $f(x) <f(y)$.
$f(x)=x^n-x^n-1$ is an increasing function on $(1,infty)$ because $f'(x)=nx^n-1-(n-1)x^n-2 >0$ ( since $x >1 >1-frac 1 n$). The required inequality simply says $x<y$ implies $f(x) <f(y)$.
answered Sep 10 at 8:40


Kavi Rama Murthy
27.1k31439
27.1k31439
add a comment |Â
add a comment |Â
up vote
3
down vote
$$1<x<yimplies 0<x-1<y-1.$$
Multiply the left inequalities $n-1$ times and the right ones once to get
$$x^n-1(x-1)<y^n-1(y-1).$$
Equality is not possible.
add a comment |Â
up vote
3
down vote
$$1<x<yimplies 0<x-1<y-1.$$
Multiply the left inequalities $n-1$ times and the right ones once to get
$$x^n-1(x-1)<y^n-1(y-1).$$
Equality is not possible.
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$$1<x<yimplies 0<x-1<y-1.$$
Multiply the left inequalities $n-1$ times and the right ones once to get
$$x^n-1(x-1)<y^n-1(y-1).$$
Equality is not possible.
$$1<x<yimplies 0<x-1<y-1.$$
Multiply the left inequalities $n-1$ times and the right ones once to get
$$x^n-1(x-1)<y^n-1(y-1).$$
Equality is not possible.
answered Sep 10 at 8:58
Yves Daoust
115k667210
115k667210
add a comment |Â
add a comment |Â
up vote
2
down vote
Need to show:
$x^n -x^n-1 lt y^n-y^n-1$;
$x^n-1(x-1) lt y^n-1(y-1)$.
Since $1<x<y$ :
$(y-1)>(x-1)>0$, and
$y^n-1 > x^n-1>0$(why?).
add a comment |Â
up vote
2
down vote
Need to show:
$x^n -x^n-1 lt y^n-y^n-1$;
$x^n-1(x-1) lt y^n-1(y-1)$.
Since $1<x<y$ :
$(y-1)>(x-1)>0$, and
$y^n-1 > x^n-1>0$(why?).
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Need to show:
$x^n -x^n-1 lt y^n-y^n-1$;
$x^n-1(x-1) lt y^n-1(y-1)$.
Since $1<x<y$ :
$(y-1)>(x-1)>0$, and
$y^n-1 > x^n-1>0$(why?).
Need to show:
$x^n -x^n-1 lt y^n-y^n-1$;
$x^n-1(x-1) lt y^n-1(y-1)$.
Since $1<x<y$ :
$(y-1)>(x-1)>0$, and
$y^n-1 > x^n-1>0$(why?).
answered Sep 10 at 8:52
Peter Szilas
8,4452617
8,4452617
add a comment |Â
add a comment |Â
up vote
2
down vote
The inequality is equivalent to
$$
x^n-1(x-1)leq y^n-1(y-1)
$$
This inequality again follows directly from the condition $1<x<y$ since all terms are positive (this is where you need the condition $x,y>1$).
The method is good, but you are on the wrong side, $1<x<y$.
– Yves Daoust
Sep 10 at 8:50
@YvesDaoust It should be correct by now.
– Janik
Sep 10 at 8:55
add a comment |Â
up vote
2
down vote
The inequality is equivalent to
$$
x^n-1(x-1)leq y^n-1(y-1)
$$
This inequality again follows directly from the condition $1<x<y$ since all terms are positive (this is where you need the condition $x,y>1$).
The method is good, but you are on the wrong side, $1<x<y$.
– Yves Daoust
Sep 10 at 8:50
@YvesDaoust It should be correct by now.
– Janik
Sep 10 at 8:55
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The inequality is equivalent to
$$
x^n-1(x-1)leq y^n-1(y-1)
$$
This inequality again follows directly from the condition $1<x<y$ since all terms are positive (this is where you need the condition $x,y>1$).
The inequality is equivalent to
$$
x^n-1(x-1)leq y^n-1(y-1)
$$
This inequality again follows directly from the condition $1<x<y$ since all terms are positive (this is where you need the condition $x,y>1$).
edited Sep 10 at 8:54
answered Sep 10 at 8:43
Janik
1,4352418
1,4352418
The method is good, but you are on the wrong side, $1<x<y$.
– Yves Daoust
Sep 10 at 8:50
@YvesDaoust It should be correct by now.
– Janik
Sep 10 at 8:55
add a comment |Â
The method is good, but you are on the wrong side, $1<x<y$.
– Yves Daoust
Sep 10 at 8:50
@YvesDaoust It should be correct by now.
– Janik
Sep 10 at 8:55
The method is good, but you are on the wrong side, $1<x<y$.
– Yves Daoust
Sep 10 at 8:50
The method is good, but you are on the wrong side, $1<x<y$.
– Yves Daoust
Sep 10 at 8:50
@YvesDaoust It should be correct by now.
– Janik
Sep 10 at 8:55
@YvesDaoust It should be correct by now.
– Janik
Sep 10 at 8:55
add a comment |Â
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Anything wrong with the question (because of the downvote)?
– EuklidAlexandria
Sep 10 at 16:06