Mixing
A binomial expansion inequality.
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Let us consider $x, y > 0$ then given $n in mathbbN$ the following inequality holds true
$$
(x+y)^frac1n leq x^frac1n + y^frac1n
$$
since
$$
((x+y)^frac1n)^n = x + y leq (x^frac1n + y^frac1n)^n = x+ y +textsome positive terms.
$$
Now let us consider $x, y$ as above and let $alpha in (0,1)$ then does the following inequality holds true, and any suggestion to prove it?
$$
(x+ y)^alpha leq x^alpha + y ^alpha.
$$
algebra-precalculus inequality arithmetic
asked Jul 31 at 14:06
JCF
18411
add a comment |Â
up vote
1
down vote
favorite
Let us consider $x, y > 0$ then given $n in mathbbN$ the following inequality holds true
$$
(x+y)^frac1n leq x^frac1n + y^frac1n
$$
since
$$
((x+y)^frac1n)^n = x + y leq (x^frac1n + y^frac1n)^n = x+ y +textsome positive terms.
$$
Now let us consider $x, y$ as above and let $alpha in (0,1)$ then does the following inequality holds true, and any suggestion to prove it?
$$
(x+ y)^alpha leq x^alpha + y ^alpha.
$$
algebra-precalculus inequality arithmetic
asked Jul 31 at 14:06
JCF
18411
Before going straight to real numbers, can you show the inequality holds for rational numbers in $(0,1)$? Once it holds for rational numbers, given any $alpha in (0,1)$, given any $varepsilon>0$, you can find $delta > 0$ such that $$left|dfracab-alpha right| < delta Longrightarrow left|(x+y)^alpha-(x+y)^a/bright| < varepsilon$$ You need to be a little careful here, but IIRC, this was a general outline of these types of proofs.
– InterstellarProbe
Jul 31 at 14:15
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let us consider $x, y > 0$ then given $n in mathbbN$ the following inequality holds true
$$
(x+y)^frac1n leq x^frac1n + y^frac1n
$$
since
$$
((x+y)^frac1n)^n = x + y leq (x^frac1n + y^frac1n)^n = x+ y +textsome positive terms.
$$
Now let us consider $x, y$ as above and let $alpha in (0,1)$ then does the following inequality holds true, and any suggestion to prove it?
$$
(x+ y)^alpha leq x^alpha + y ^alpha.
$$
algebra-precalculus inequality arithmetic
asked Jul 31 at 14:06
JCF
18411
Let us consider $x, y > 0$ then given $n in mathbbN$ the following inequality holds true
$$
(x+y)^frac1n leq x^frac1n + y^frac1n
$$
since
$$
((x+y)^frac1n)^n = x + y leq (x^frac1n + y^frac1n)^n = x+ y +textsome positive terms.
$$
Now let us consider $x, y$ as above and let $alpha in (0,1)$ then does the following inequality holds true, and any suggestion to prove it?
$$
(x+ y)^alpha leq x^alpha + y ^alpha.
$$
algebra-precalculus inequality arithmetic
asked Jul 31 at 14:06
JCF
18411
asked Jul 31 at 14:06
JCF
18411
asked Jul 31 at 14:06
JCF
18411
asked Jul 31 at 14:06
JCF
18411
18411
Before going straight to real numbers, can you show the inequality holds for rational numbers in $(0,1)$? Once it holds for rational numbers, given any $alpha in (0,1)$, given any $varepsilon>0$, you can find $delta > 0$ such that $$left|dfracab-alpha right| < delta Longrightarrow left|(x+y)^alpha-(x+y)^a/bright| < varepsilon$$ You need to be a little careful here, but IIRC, this was a general outline of these types of proofs.
– InterstellarProbe
Jul 31 at 14:15
add a comment |Â
Before going straight to real numbers, can you show the inequality holds for rational numbers in $(0,1)$? Once it holds for rational numbers, given any $alpha in (0,1)$, given any $varepsilon>0$, you can find $delta > 0$ such that $$left|dfracab-alpha right| < delta Longrightarrow left|(x+y)^alpha-(x+y)^a/bright| < varepsilon$$ You need to be a little careful here, but IIRC, this was a general outline of these types of proofs.
– InterstellarProbe
Jul 31 at 14:15
Before going straight to real numbers, can you show the inequality holds for rational numbers in $(0,1)$? Once it holds for rational numbers, given any $alpha in (0,1)$, given any $varepsilon>0$, you can find $delta > 0$ such that $$left|dfracab-alpha right| < delta Longrightarrow left|(x+y)^alpha-(x+y)^a/bright| < varepsilon$$ You need to be a little careful here, but IIRC, this was a general outline of these types of proofs.
– InterstellarProbe
Jul 31 at 14:15
Before going straight to real numbers, can you show the inequality holds for rational numbers in $(0,1)$? Once it holds for rational numbers, given any $alpha in (0,1)$, given any $varepsilon>0$, you can find $delta > 0$ such that $$left|dfracab-alpha right| < delta Longrightarrow left|(x+y)^alpha-(x+y)^a/bright| < varepsilon$$ You need to be a little careful here, but IIRC, this was a general outline of these types of proofs.
– InterstellarProbe
Jul 31 at 14:15
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
$$
(x+y)^alpha-x^alpha=int_x^x+yalpha t^alpha-1,dtle int_x^x+yalpha (t-x)^alpha-1,dt=y^alpha-requirecancelcancel0^alpha.
$$
answered Jul 31 at 14:17
Mike Earnest
14.7k11644
I wish I was clever like that. (Although you could use some dt's on those integrals--smile).
– Leonard Blackburn
Jul 31 at 14:27
@LeonardBlackburn To be fair, I remembered the solution from an analysis homework way back.
– Mike Earnest
Jul 31 at 14:34
Thanks... it seems that I always find an analysis homework that I haven't done
– JCF
Jul 31 at 14:46
add a comment |Â
up vote
0
down vote
Also, you can use Karamata here.
Indeed, let $f(x)=x^alpha$ and $xgeq y$.
Thus, $f$ is a concave function and $(x+y,0)succ(x,y).$
Id est, by Kartamata
$$f(x+y)+f(0)leq f(x)+f(y),$$ which is your inequality.
answered Jul 31 at 19:26
Michael Rozenberg
87.5k1577179
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$$
(x+y)^alpha-x^alpha=int_x^x+yalpha t^alpha-1,dtle int_x^x+yalpha (t-x)^alpha-1,dt=y^alpha-requirecancelcancel0^alpha.
$$
answered Jul 31 at 14:17
Mike Earnest
14.7k11644
I wish I was clever like that. (Although you could use some dt's on those integrals--smile).
– Leonard Blackburn
Jul 31 at 14:27
@LeonardBlackburn To be fair, I remembered the solution from an analysis homework way back.
– Mike Earnest
Jul 31 at 14:34
Thanks... it seems that I always find an analysis homework that I haven't done
– JCF
Jul 31 at 14:46
add a comment |Â
up vote
3
down vote
accepted
$$
(x+y)^alpha-x^alpha=int_x^x+yalpha t^alpha-1,dtle int_x^x+yalpha (t-x)^alpha-1,dt=y^alpha-requirecancelcancel0^alpha.
$$
answered Jul 31 at 14:17
Mike Earnest
14.7k11644
I wish I was clever like that. (Although you could use some dt's on those integrals--smile).
– Leonard Blackburn
Jul 31 at 14:27
@LeonardBlackburn To be fair, I remembered the solution from an analysis homework way back.
– Mike Earnest
Jul 31 at 14:34
Thanks... it seems that I always find an analysis homework that I haven't done
– JCF
Jul 31 at 14:46
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$$
(x+y)^alpha-x^alpha=int_x^x+yalpha t^alpha-1,dtle int_x^x+yalpha (t-x)^alpha-1,dt=y^alpha-requirecancelcancel0^alpha.
$$
answered Jul 31 at 14:17
Mike Earnest
14.7k11644
$$
(x+y)^alpha-x^alpha=int_x^x+yalpha t^alpha-1,dtle int_x^x+yalpha (t-x)^alpha-1,dt=y^alpha-requirecancelcancel0^alpha.
$$
answered Jul 31 at 14:17
Mike Earnest
14.7k11644
edited Jul 31 at 14:33
answered Jul 31 at 14:17
Mike Earnest
14.7k11644
answered Jul 31 at 14:17
Mike Earnest
14.7k11644
answered Jul 31 at 14:17
Mike Earnest
14.7k11644
14.7k11644
I wish I was clever like that. (Although you could use some dt's on those integrals--smile).
– Leonard Blackburn
Jul 31 at 14:27
@LeonardBlackburn To be fair, I remembered the solution from an analysis homework way back.
– Mike Earnest
Jul 31 at 14:34
Thanks... it seems that I always find an analysis homework that I haven't done
– JCF
Jul 31 at 14:46
add a comment |Â
I wish I was clever like that. (Although you could use some dt's on those integrals--smile).
– Leonard Blackburn
Jul 31 at 14:27
@LeonardBlackburn To be fair, I remembered the solution from an analysis homework way back.
– Mike Earnest
Jul 31 at 14:34
Thanks... it seems that I always find an analysis homework that I haven't done
– JCF
Jul 31 at 14:46
I wish I was clever like that. (Although you could use some dt's on those integrals--smile).
– Leonard Blackburn
Jul 31 at 14:27
I wish I was clever like that. (Although you could use some dt's on those integrals--smile).
– Leonard Blackburn
Jul 31 at 14:27
@LeonardBlackburn To be fair, I remembered the solution from an analysis homework way back.
– Mike Earnest
Jul 31 at 14:34
@LeonardBlackburn To be fair, I remembered the solution from an analysis homework way back.
– Mike Earnest
Jul 31 at 14:34
Thanks... it seems that I always find an analysis homework that I haven't done
– JCF
Jul 31 at 14:46
Thanks... it seems that I always find an analysis homework that I haven't done
– JCF
Jul 31 at 14:46
add a comment |Â
up vote
0
down vote
Also, you can use Karamata here.
Indeed, let $f(x)=x^alpha$ and $xgeq y$.
Thus, $f$ is a concave function and $(x+y,0)succ(x,y).$
Id est, by Kartamata
$$f(x+y)+f(0)leq f(x)+f(y),$$ which is your inequality.
answered Jul 31 at 19:26
Michael Rozenberg
87.5k1577179
add a comment |Â
up vote
0
down vote
Also, you can use Karamata here.
Indeed, let $f(x)=x^alpha$ and $xgeq y$.
Thus, $f$ is a concave function and $(x+y,0)succ(x,y).$
Id est, by Kartamata
$$f(x+y)+f(0)leq f(x)+f(y),$$ which is your inequality.
answered Jul 31 at 19:26
Michael Rozenberg
87.5k1577179
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Also, you can use Karamata here.
Indeed, let $f(x)=x^alpha$ and $xgeq y$.
Thus, $f$ is a concave function and $(x+y,0)succ(x,y).$
Id est, by Kartamata
$$f(x+y)+f(0)leq f(x)+f(y),$$ which is your inequality.
answered Jul 31 at 19:26
Michael Rozenberg
87.5k1577179
Also, you can use Karamata here.
Indeed, let $f(x)=x^alpha$ and $xgeq y$.
Thus, $f$ is a concave function and $(x+y,0)succ(x,y).$
Id est, by Kartamata
$$f(x+y)+f(0)leq f(x)+f(y),$$ which is your inequality.
answered Jul 31 at 19:26
Michael Rozenberg
87.5k1577179
edited Jul 31 at 19:36
answered Jul 31 at 19:26
Michael Rozenberg
87.5k1577179
answered Jul 31 at 19:26
Michael Rozenberg
87.5k1577179
answered Jul 31 at 19:26
Michael Rozenberg
87.5k1577179
87.5k1577179
add a comment |Â
add a comment |Â
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Before going straight to real numbers, can you show the inequality holds for rational numbers in $(0,1)$? Once it holds for rational numbers, given any $alpha in (0,1)$, given any $varepsilon>0$, you can find $delta > 0$ such that $$left|dfracab-alpha right| < delta Longrightarrow left|(x+y)^alpha-(x+y)^a/bright| < varepsilon$$ You need to be a little careful here, but IIRC, this was a general outline of these types of proofs.
– InterstellarProbe
Jul 31 at 14:15