$x,y in R$ and $x neq y$ Show that $e^frac{x+y}2 lt frac12 (e^x + e^y)$
$begingroup$
Let $x,y in R$ and $x neq y$. Show that $e^frac{x+y}2 lt frac12 (e^x + e^y)$
Struggling with this proof atm.
Tried every approach I could think of. Attempted to look at it written as a series, as a limit and tried playing around with the general rules for exponential functions.
calculus inequality
edited Nov 29 '18 at 18:23
Richard
3801211
asked Nov 29 '18 at 17:58
NhefNhef
91
$endgroup$
add a comment |
$begingroup$
Let $x,y in R$ and $x neq y$. Show that $e^frac{x+y}2 lt frac12 (e^x + e^y)$
Struggling with this proof atm.
Tried every approach I could think of. Attempted to look at it written as a series, as a limit and tried playing around with the general rules for exponential functions.
calculus inequality
edited Nov 29 '18 at 18:23
Richard
3801211
asked Nov 29 '18 at 17:58
NhefNhef
91
$endgroup$
1
$begingroup$
This is just the arithmetic-geometric mean inequality.
$endgroup$
– Wojowu
Nov 29 '18 at 18:03
$begingroup$
Can people please stop tagging real analysis to whatever has to do with calculus ?
$endgroup$
– Rebellos
Nov 29 '18 at 18:05
add a comment |
$begingroup$
Let $x,y in R$ and $x neq y$. Show that $e^frac{x+y}2 lt frac12 (e^x + e^y)$
Struggling with this proof atm.
Tried every approach I could think of. Attempted to look at it written as a series, as a limit and tried playing around with the general rules for exponential functions.
calculus inequality
edited Nov 29 '18 at 18:23
Richard
3801211
asked Nov 29 '18 at 17:58
NhefNhef
91
$endgroup$
Let $x,y in R$ and $x neq y$. Show that $e^frac{x+y}2 lt frac12 (e^x + e^y)$
Struggling with this proof atm.
Tried every approach I could think of. Attempted to look at it written as a series, as a limit and tried playing around with the general rules for exponential functions.
calculus inequality
calculus inequality
edited Nov 29 '18 at 18:23
Richard
3801211
asked Nov 29 '18 at 17:58
NhefNhef
91
edited Nov 29 '18 at 18:23
Richard
3801211
asked Nov 29 '18 at 17:58
NhefNhef
91
edited Nov 29 '18 at 18:23
Richard
3801211
edited Nov 29 '18 at 18:23
Richard
3801211
edited Nov 29 '18 at 18:23
Richard
3801211
3801211
asked Nov 29 '18 at 17:58
NhefNhef
91
asked Nov 29 '18 at 17:58
NhefNhef
91
asked Nov 29 '18 at 17:58
NhefNhef
91
91
1
$begingroup$
This is just the arithmetic-geometric mean inequality.
$endgroup$
– Wojowu
Nov 29 '18 at 18:03
$begingroup$
Can people please stop tagging real analysis to whatever has to do with calculus ?
$endgroup$
– Rebellos
Nov 29 '18 at 18:05
add a comment |
1
$begingroup$
This is just the arithmetic-geometric mean inequality.
$endgroup$
– Wojowu
Nov 29 '18 at 18:03
$begingroup$
Can people please stop tagging real analysis to whatever has to do with calculus ?
$endgroup$
– Rebellos
Nov 29 '18 at 18:05
1
1
$begingroup$
This is just the arithmetic-geometric mean inequality.
$endgroup$
– Wojowu
Nov 29 '18 at 18:03
$begingroup$
This is just the arithmetic-geometric mean inequality.
$endgroup$
– Wojowu
Nov 29 '18 at 18:03
$begingroup$
Can people please stop tagging real analysis to whatever has to do with calculus ?
$endgroup$
– Rebellos
Nov 29 '18 at 18:05
$begingroup$
Can people please stop tagging real analysis to whatever has to do with calculus ?
$endgroup$
– Rebellos
Nov 29 '18 at 18:05
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
if $ane b$
$(a-b)^2 > 0\
a^2 + b^2 > 2ab\
ab < frac 12 (a^2 + b^2)$
This is the AM-GM inequality.
$a = e^frac x2, b= e^frac y2$
answered Nov 29 '18 at 18:03
Doug MDoug M
44.8k31854
$endgroup$
add a comment |
$begingroup$
By convexity for $f(x)=e^x$, that is by Jensen's inequality for $xneq y$
$$f(lambda x+(1-lambda)y) < lambda f(x)+(1-lambda)f(y)$$
with $lambda=frac12$ we have
$$e^{frac{x+y}2} < frac12 (e^x + e^y)$$
answered Nov 29 '18 at 18:16
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
We have $${(e^{xover 2}-e^{yover 2})^2>0\e^x+e^y>2e^{xover 2}e^{yover 2}\{1over 2}(e^x+e^y)>e^{x+yover 2}}$$
answered Nov 29 '18 at 18:21
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
if $ane b$
$(a-b)^2 > 0\
a^2 + b^2 > 2ab\
ab < frac 12 (a^2 + b^2)$
This is the AM-GM inequality.
$a = e^frac x2, b= e^frac y2$
answered Nov 29 '18 at 18:03
Doug MDoug M
44.8k31854
$endgroup$
add a comment |
$begingroup$
if $ane b$
$(a-b)^2 > 0\
a^2 + b^2 > 2ab\
ab < frac 12 (a^2 + b^2)$
This is the AM-GM inequality.
$a = e^frac x2, b= e^frac y2$
answered Nov 29 '18 at 18:03
Doug MDoug M
44.8k31854
$endgroup$
add a comment |
$begingroup$
if $ane b$
$(a-b)^2 > 0\
a^2 + b^2 > 2ab\
ab < frac 12 (a^2 + b^2)$
This is the AM-GM inequality.
$a = e^frac x2, b= e^frac y2$
answered Nov 29 '18 at 18:03
Doug MDoug M
44.8k31854
$endgroup$
if $ane b$
$(a-b)^2 > 0\
a^2 + b^2 > 2ab\
ab < frac 12 (a^2 + b^2)$
This is the AM-GM inequality.
$a = e^frac x2, b= e^frac y2$
answered Nov 29 '18 at 18:03
Doug MDoug M
44.8k31854
answered Nov 29 '18 at 18:03
Doug MDoug M
44.8k31854
answered Nov 29 '18 at 18:03
Doug MDoug M
44.8k31854
answered Nov 29 '18 at 18:03
Doug MDoug M
44.8k31854
44.8k31854
add a comment |
add a comment |
$begingroup$
By convexity for $f(x)=e^x$, that is by Jensen's inequality for $xneq y$
$$f(lambda x+(1-lambda)y) < lambda f(x)+(1-lambda)f(y)$$
with $lambda=frac12$ we have
$$e^{frac{x+y}2} < frac12 (e^x + e^y)$$
answered Nov 29 '18 at 18:16
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
By convexity for $f(x)=e^x$, that is by Jensen's inequality for $xneq y$
$$f(lambda x+(1-lambda)y) < lambda f(x)+(1-lambda)f(y)$$
with $lambda=frac12$ we have
$$e^{frac{x+y}2} < frac12 (e^x + e^y)$$
answered Nov 29 '18 at 18:16
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
By convexity for $f(x)=e^x$, that is by Jensen's inequality for $xneq y$
$$f(lambda x+(1-lambda)y) < lambda f(x)+(1-lambda)f(y)$$
with $lambda=frac12$ we have
$$e^{frac{x+y}2} < frac12 (e^x + e^y)$$
answered Nov 29 '18 at 18:16
gimusigimusi
92.8k84494
$endgroup$
By convexity for $f(x)=e^x$, that is by Jensen's inequality for $xneq y$
$$f(lambda x+(1-lambda)y) < lambda f(x)+(1-lambda)f(y)$$
with $lambda=frac12$ we have
$$e^{frac{x+y}2} < frac12 (e^x + e^y)$$
answered Nov 29 '18 at 18:16
gimusigimusi
92.8k84494
answered Nov 29 '18 at 18:16
gimusigimusi
92.8k84494
answered Nov 29 '18 at 18:16
gimusigimusi
92.8k84494
answered Nov 29 '18 at 18:16
gimusigimusi
92.8k84494
92.8k84494
add a comment |
add a comment |
$begingroup$
We have $${(e^{xover 2}-e^{yover 2})^2>0\e^x+e^y>2e^{xover 2}e^{yover 2}\{1over 2}(e^x+e^y)>e^{x+yover 2}}$$
answered Nov 29 '18 at 18:21
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
add a comment |
$begingroup$
We have $${(e^{xover 2}-e^{yover 2})^2>0\e^x+e^y>2e^{xover 2}e^{yover 2}\{1over 2}(e^x+e^y)>e^{x+yover 2}}$$
answered Nov 29 '18 at 18:21
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
add a comment |
$begingroup$
We have $${(e^{xover 2}-e^{yover 2})^2>0\e^x+e^y>2e^{xover 2}e^{yover 2}\{1over 2}(e^x+e^y)>e^{x+yover 2}}$$
answered Nov 29 '18 at 18:21
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
We have $${(e^{xover 2}-e^{yover 2})^2>0\e^x+e^y>2e^{xover 2}e^{yover 2}\{1over 2}(e^x+e^y)>e^{x+yover 2}}$$
answered Nov 29 '18 at 18:21
Mostafa AyazMostafa Ayaz
15.5k3939
answered Nov 29 '18 at 18:21
Mostafa AyazMostafa Ayaz
15.5k3939
answered Nov 29 '18 at 18:21
Mostafa AyazMostafa Ayaz
15.5k3939
answered Nov 29 '18 at 18:21
Mostafa AyazMostafa Ayaz
15.5k3939
15.5k3939
add a comment |
add a comment |
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1
$begingroup$
This is just the arithmetic-geometric mean inequality.
$endgroup$
– Wojowu
Nov 29 '18 at 18:03
$begingroup$
Can people please stop tagging real analysis to whatever has to do with calculus ?
$endgroup$
– Rebellos
Nov 29 '18 at 18:05