Mixing
Proof for $a^c + b^c > (a + b)^c$ when $0 < c 0$ and $1/c$ is nonintegral
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What is the proof for the statement $a^c + b^c > (a + b)^c$ when $0 < c < 1$, $a, b> 0$ and $1/c$ is non-integral? I have a very simple proof for this statement when $1/c$ is an integer (namely, just raise both sides to $1/c$ and the proof immediately follows from binomial expansion of $(a^c + b^c)^{1/c}$).
But what about the case when $1/c$ is non-integral?
algebra-precalculus inequality
edited Dec 30 '18 at 20:20
Eric Wofsey
181k12208336
asked Dec 30 '18 at 4:47
user1246462user1246462
1503
$endgroup$
add a comment |
$begingroup$
What is the proof for the statement $a^c + b^c > (a + b)^c$ when $0 < c < 1$, $a, b> 0$ and $1/c$ is non-integral? I have a very simple proof for this statement when $1/c$ is an integer (namely, just raise both sides to $1/c$ and the proof immediately follows from binomial expansion of $(a^c + b^c)^{1/c}$).
But what about the case when $1/c$ is non-integral?
algebra-precalculus inequality
edited Dec 30 '18 at 20:20
Eric Wofsey
181k12208336
asked Dec 30 '18 at 4:47
user1246462user1246462
1503
$endgroup$
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Start with $$dfrac ab=d$$
$endgroup$
– lab bhattacharjee
Dec 30 '18 at 4:51
2
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Is Jensen's inequality available to you?
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– DanielV
Dec 30 '18 at 4:54
3
$begingroup$
This is just a special case of $|cdot|_{ell_q}<|cdot|_{ell_p}$ for $0<p<q<infty$. See here: math.stackexchange.com/questions/4094/…
$endgroup$
– Ben W
Dec 30 '18 at 4:56
add a comment |
$begingroup$
What is the proof for the statement $a^c + b^c > (a + b)^c$ when $0 < c < 1$, $a, b> 0$ and $1/c$ is non-integral? I have a very simple proof for this statement when $1/c$ is an integer (namely, just raise both sides to $1/c$ and the proof immediately follows from binomial expansion of $(a^c + b^c)^{1/c}$).
But what about the case when $1/c$ is non-integral?
algebra-precalculus inequality
edited Dec 30 '18 at 20:20
Eric Wofsey
181k12208336
asked Dec 30 '18 at 4:47
user1246462user1246462
1503
$endgroup$
What is the proof for the statement $a^c + b^c > (a + b)^c$ when $0 < c < 1$, $a, b> 0$ and $1/c$ is non-integral? I have a very simple proof for this statement when $1/c$ is an integer (namely, just raise both sides to $1/c$ and the proof immediately follows from binomial expansion of $(a^c + b^c)^{1/c}$).
But what about the case when $1/c$ is non-integral?
algebra-precalculus inequality
algebra-precalculus inequality
edited Dec 30 '18 at 20:20
Eric Wofsey
181k12208336
asked Dec 30 '18 at 4:47
user1246462user1246462
1503
edited Dec 30 '18 at 20:20
Eric Wofsey
181k12208336
asked Dec 30 '18 at 4:47
user1246462user1246462
1503
edited Dec 30 '18 at 20:20
Eric Wofsey
181k12208336
edited Dec 30 '18 at 20:20
Eric Wofsey
181k12208336
edited Dec 30 '18 at 20:20
Eric Wofsey
181k12208336
181k12208336
asked Dec 30 '18 at 4:47
user1246462user1246462
1503
asked Dec 30 '18 at 4:47
user1246462user1246462
1503
asked Dec 30 '18 at 4:47
user1246462user1246462
1503
1503
$begingroup$
Start with $$dfrac ab=d$$
$endgroup$
– lab bhattacharjee
Dec 30 '18 at 4:51
2
$begingroup$
Is Jensen's inequality available to you?
$endgroup$
– DanielV
Dec 30 '18 at 4:54
3
$begingroup$
This is just a special case of $|cdot|_{ell_q}<|cdot|_{ell_p}$ for $0<p<q<infty$. See here: math.stackexchange.com/questions/4094/…
$endgroup$
– Ben W
Dec 30 '18 at 4:56
add a comment |
$begingroup$
Start with $$dfrac ab=d$$
$endgroup$
– lab bhattacharjee
Dec 30 '18 at 4:51
2
$begingroup$
Is Jensen's inequality available to you?
$endgroup$
– DanielV
Dec 30 '18 at 4:54
3
$begingroup$
This is just a special case of $|cdot|_{ell_q}<|cdot|_{ell_p}$ for $0<p<q<infty$. See here: math.stackexchange.com/questions/4094/…
$endgroup$
– Ben W
Dec 30 '18 at 4:56
$begingroup$
Start with $$dfrac ab=d$$
$endgroup$
– lab bhattacharjee
Dec 30 '18 at 4:51
$begingroup$
Start with $$dfrac ab=d$$
$endgroup$
– lab bhattacharjee
Dec 30 '18 at 4:51
2
2
$begingroup$
Is Jensen's inequality available to you?
$endgroup$
– DanielV
Dec 30 '18 at 4:54
$begingroup$
Is Jensen's inequality available to you?
$endgroup$
– DanielV
Dec 30 '18 at 4:54
3
3
$begingroup$
This is just a special case of $|cdot|_{ell_q}<|cdot|_{ell_p}$ for $0<p<q<infty$. See here: math.stackexchange.com/questions/4094/…
$endgroup$
– Ben W
Dec 30 '18 at 4:56
$begingroup$
This is just a special case of $|cdot|_{ell_q}<|cdot|_{ell_p}$ for $0<p<q<infty$. See here: math.stackexchange.com/questions/4094/…
$endgroup$
– Ben W
Dec 30 '18 at 4:56
add a comment |
2 Answers
2
active
oldest
votes
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$(1+t)^{c}-1-t^{c}$ vanishes at $t=0$ and is decreasing in $t>0$ (because its derivative is negative). Hence $(1+t)^{c}<1+t^{c}$. Put $t=frac b a$ and muliply both sides by $a^{c}$.
answered Dec 30 '18 at 5:21
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
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add a comment |
$begingroup$
The function $f:(0,infty)to mathbb{R}$ defined by $f(x)=x^c$ with $0<c<1$ is strictly concave. Therefore $$f(a)+f(b)> f(a+b).$$
answered Jan 1 at 19:55
user376343user376343
3,3032825
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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$begingroup$
$(1+t)^{c}-1-t^{c}$ vanishes at $t=0$ and is decreasing in $t>0$ (because its derivative is negative). Hence $(1+t)^{c}<1+t^{c}$. Put $t=frac b a$ and muliply both sides by $a^{c}$.
answered Dec 30 '18 at 5:21
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
$endgroup$
add a comment |
$begingroup$
$(1+t)^{c}-1-t^{c}$ vanishes at $t=0$ and is decreasing in $t>0$ (because its derivative is negative). Hence $(1+t)^{c}<1+t^{c}$. Put $t=frac b a$ and muliply both sides by $a^{c}$.
answered Dec 30 '18 at 5:21
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
$endgroup$
add a comment |
$begingroup$
$(1+t)^{c}-1-t^{c}$ vanishes at $t=0$ and is decreasing in $t>0$ (because its derivative is negative). Hence $(1+t)^{c}<1+t^{c}$. Put $t=frac b a$ and muliply both sides by $a^{c}$.
answered Dec 30 '18 at 5:21
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
$endgroup$
$(1+t)^{c}-1-t^{c}$ vanishes at $t=0$ and is decreasing in $t>0$ (because its derivative is negative). Hence $(1+t)^{c}<1+t^{c}$. Put $t=frac b a$ and muliply both sides by $a^{c}$.
answered Dec 30 '18 at 5:21
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
answered Dec 30 '18 at 5:21
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
answered Dec 30 '18 at 5:21
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
answered Dec 30 '18 at 5:21
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
52.8k32055
add a comment |
add a comment |
$begingroup$
The function $f:(0,infty)to mathbb{R}$ defined by $f(x)=x^c$ with $0<c<1$ is strictly concave. Therefore $$f(a)+f(b)> f(a+b).$$
answered Jan 1 at 19:55
user376343user376343
3,3032825
$endgroup$
add a comment |
$begingroup$
The function $f:(0,infty)to mathbb{R}$ defined by $f(x)=x^c$ with $0<c<1$ is strictly concave. Therefore $$f(a)+f(b)> f(a+b).$$
answered Jan 1 at 19:55
user376343user376343
3,3032825
$endgroup$
add a comment |
$begingroup$
The function $f:(0,infty)to mathbb{R}$ defined by $f(x)=x^c$ with $0<c<1$ is strictly concave. Therefore $$f(a)+f(b)> f(a+b).$$
answered Jan 1 at 19:55
user376343user376343
3,3032825
$endgroup$
The function $f:(0,infty)to mathbb{R}$ defined by $f(x)=x^c$ with $0<c<1$ is strictly concave. Therefore $$f(a)+f(b)> f(a+b).$$
answered Jan 1 at 19:55
user376343user376343
3,3032825
answered Jan 1 at 19:55
user376343user376343
3,3032825
answered Jan 1 at 19:55
user376343user376343
3,3032825
answered Jan 1 at 19:55
user376343user376343
3,3032825
3,3032825
add a comment |
add a comment |
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$begingroup$
Start with $$dfrac ab=d$$
$endgroup$
– lab bhattacharjee
Dec 30 '18 at 4:51
2
$begingroup$
Is Jensen's inequality available to you?
$endgroup$
– DanielV
Dec 30 '18 at 4:54
3
$begingroup$
This is just a special case of $|cdot|_{ell_q}<|cdot|_{ell_p}$ for $0<p<q<infty$. See here: math.stackexchange.com/questions/4094/…
$endgroup$
– Ben W
Dec 30 '18 at 4:56