Mixing
Inequality with exponents $x^x+y^y ge x^y +y^x$
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Let $x,y$ be positive numbers. Prove that $x^x+y^y ge x^y +y^x$.
This question appeared in Summer 1991 Russian Olympiad team test. Apparently, I tried to come up with different approach such as Jensen, Karamata's inequality and nothing works so far. I just need a discussion here. Hints are not necessary.
inequality contest-math exponentiation
edited 2 days ago
Michael Rozenberg
94.2k1588183
asked Jun 4 '13 at 10:52
Viet Hoang Quoc
544
add a comment |
up vote
7
down vote
favorite
Let $x,y$ be positive numbers. Prove that $x^x+y^y ge x^y +y^x$.
This question appeared in Summer 1991 Russian Olympiad team test. Apparently, I tried to come up with different approach such as Jensen, Karamata's inequality and nothing works so far. I just need a discussion here. Hints are not necessary.
inequality contest-math exponentiation
edited 2 days ago
Michael Rozenberg
94.2k1588183
asked Jun 4 '13 at 10:52
Viet Hoang Quoc
544
:use $AM-GM$ inequality or define function $(f(x)=x^x)$
– M.H
Jun 4 '13 at 11:03
Can you elaborate a bit more because I have used AM-GM but the difference in the power could be a major issue?
– Viet Hoang Quoc
Jun 4 '13 at 11:13
There is a long ML thread about this and related ones at artofproblemsolving.com/Forum/viewtopic.php?f=52&t=118722
– Wolfgang
Oct 12 '13 at 16:41
add a comment |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Let $x,y$ be positive numbers. Prove that $x^x+y^y ge x^y +y^x$.
This question appeared in Summer 1991 Russian Olympiad team test. Apparently, I tried to come up with different approach such as Jensen, Karamata's inequality and nothing works so far. I just need a discussion here. Hints are not necessary.
inequality contest-math exponentiation
edited 2 days ago
Michael Rozenberg
94.2k1588183
asked Jun 4 '13 at 10:52
Viet Hoang Quoc
544
Let $x,y$ be positive numbers. Prove that $x^x+y^y ge x^y +y^x$.
This question appeared in Summer 1991 Russian Olympiad team test. Apparently, I tried to come up with different approach such as Jensen, Karamata's inequality and nothing works so far. I just need a discussion here. Hints are not necessary.
inequality contest-math exponentiation
inequality contest-math exponentiation
edited 2 days ago
Michael Rozenberg
94.2k1588183
asked Jun 4 '13 at 10:52
Viet Hoang Quoc
544
edited 2 days ago
Michael Rozenberg
94.2k1588183
asked Jun 4 '13 at 10:52
Viet Hoang Quoc
544
edited 2 days ago
Michael Rozenberg
94.2k1588183
edited 2 days ago
Michael Rozenberg
94.2k1588183
edited 2 days ago
Michael Rozenberg
94.2k1588183
94.2k1588183
asked Jun 4 '13 at 10:52
Viet Hoang Quoc
544
asked Jun 4 '13 at 10:52
Viet Hoang Quoc
544
asked Jun 4 '13 at 10:52
Viet Hoang Quoc
544
544
:use $AM-GM$ inequality or define function $(f(x)=x^x)$
– M.H
Jun 4 '13 at 11:03
Can you elaborate a bit more because I have used AM-GM but the difference in the power could be a major issue?
– Viet Hoang Quoc
Jun 4 '13 at 11:13
There is a long ML thread about this and related ones at artofproblemsolving.com/Forum/viewtopic.php?f=52&t=118722
– Wolfgang
Oct 12 '13 at 16:41
add a comment |
:use $AM-GM$ inequality or define function $(f(x)=x^x)$
– M.H
Jun 4 '13 at 11:03
Can you elaborate a bit more because I have used AM-GM but the difference in the power could be a major issue?
– Viet Hoang Quoc
Jun 4 '13 at 11:13
There is a long ML thread about this and related ones at artofproblemsolving.com/Forum/viewtopic.php?f=52&t=118722
– Wolfgang
Oct 12 '13 at 16:41
:use $AM-GM$ inequality or define function $(f(x)=x^x)$
– M.H
Jun 4 '13 at 11:03
:use $AM-GM$ inequality or define function $(f(x)=x^x)$
– M.H
Jun 4 '13 at 11:03
Can you elaborate a bit more because I have used AM-GM but the difference in the power could be a major issue?
– Viet Hoang Quoc
Jun 4 '13 at 11:13
Can you elaborate a bit more because I have used AM-GM but the difference in the power could be a major issue?
– Viet Hoang Quoc
Jun 4 '13 at 11:13
There is a long ML thread about this and related ones at artofproblemsolving.com/Forum/viewtopic.php?f=52&t=118722
– Wolfgang
Oct 12 '13 at 16:41
There is a long ML thread about this and related ones at artofproblemsolving.com/Forum/viewtopic.php?f=52&t=118722
– Wolfgang
Oct 12 '13 at 16:41
add a comment |
4 Answers
4
active
oldest
votes
up vote
1
down vote
without loss of generality we only prove by $0le yle xle 1$,let
$$f(a)=a^{bx}-a^{by},xge age y,1ge bx-byge 0$$
then
$$(a^{bx-by})'_{a}=dfrac{bx-by}{a}cdot a^{bx-by}>0,Longrightarrow a^{bx-by}ge y^{bx-by}cdots (1)$$
since use $AM-GM$ inequality,we have
$$y^{1+y-x}1^{x+xy-y^2-y}leleft(dfrac{x}{1+xy-y^2}right)^{1+xy-y^2}le xcdots (2)$$
and
$$f'(a)=dfrac{bx}{a}cdot a^{bx}-dfrac{by}{a}cdot a^{by}=dfrac{ba^{by}}{a}(xa^{b(x-y)}-y)=$$
so use $(1),(2)$ we have
$$f'(a)ge ba^{by-1}(xy^{b(x-y)}-y)=ba^{by-1}y^{b(x-y)}(x-y^{1-b(x-y)})ge 0
$$
so
$$f(x)ge f(y)Longrightarrow x^{bx}+y^{by}ge x^{by}+y^{bx}$$
let $b=1$
we have
$$x^x+y^yge x^y+y^x$$
answered Jun 4 '13 at 12:21
math110
32.4k455215
The differentiation step does not look right to me, we have either $ (a^x)' = ln (a) times a^x $ or $(x^a)'=a times x^{a-1}$.
– Viet Hoang Quoc
Jun 5 '13 at 8:15
I use $(x^a)'_{x}=atimes x^{a-1}$..
– math110
Sep 3 '13 at 0:43
add a comment |
up vote
1
down vote
Let $xgeq y$.
We'll consider two cases.
$xgeq1$.
Hence, $$x^x+y^y=x^yleft(x^{x-y}-1right)+x^y+y^ygeq y^yleft(x^{x-y}-1right)+x^y+y^ygeq$$
$$geq y^yleft(y^{x-y}-1right)+x^y+y^ygeq x^y+y^x.$$
2. $1>xgeq y>0.$
Let $f(t)=t^x-t^y$, where $tin[y,1).$
Thus, by Bernoulli we obtain:
$$f'(t)=xt^{x-1}-yt^{y-1}=t^{y-1}left(xt^{x-y}-yright)geq$$
$$geq t^{y-1}left(xy^{x-y}-yright)=t^{y-1}y^{x-y}left(x-y^{1-x+y}right)=$$
$$=t^{y-1}y^{x-y}left(x-(1+(y-1))^{1-x+y}right)geq$$
$$geq t^{y-1}y^{x-y}left(x-(1+(y-1)(1-x+y))right)=t^{y-1}y^{x-y}y(x-y)geq0,$$
which gives $$f(x)geq f(y)$$ or
$$x^x-x^ygeq y^x-y^y$$ or
$$x^x+y^ygeq x^y+y^x$$ and we are done!
answered 2 days ago
Michael Rozenberg
94.2k1588183
add a comment |
up vote
0
down vote
Assume $xge y$ by symmetry. We want
$$x^x-x^yge y^x-y^y.$$
i.e. $$x^y(x^{x-y}-1)ge y^y(y^{x-y}-1),$$
which is then easy to show by listing all possible cases according to how $x,y$ compare to 1.
answered Jun 4 '13 at 11:38
Spook
2,2121033
add a comment |
up vote
-3
down vote
Use induction, taking the base case x=0
answered Jun 4 '13 at 11:21
stancamp1
1
3
This happens for all positive real number and induction clearly does not work nicely here.
– Viet Hoang Quoc
Jun 4 '13 at 11:29
2
This is wrong in the general case. @stancamp1, you may want to edit this answer or delete it in order to avoid downvotes.
– DonAntonio
Jun 4 '13 at 11:39
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
without loss of generality we only prove by $0le yle xle 1$,let
$$f(a)=a^{bx}-a^{by},xge age y,1ge bx-byge 0$$
then
$$(a^{bx-by})'_{a}=dfrac{bx-by}{a}cdot a^{bx-by}>0,Longrightarrow a^{bx-by}ge y^{bx-by}cdots (1)$$
since use $AM-GM$ inequality,we have
$$y^{1+y-x}1^{x+xy-y^2-y}leleft(dfrac{x}{1+xy-y^2}right)^{1+xy-y^2}le xcdots (2)$$
and
$$f'(a)=dfrac{bx}{a}cdot a^{bx}-dfrac{by}{a}cdot a^{by}=dfrac{ba^{by}}{a}(xa^{b(x-y)}-y)=$$
so use $(1),(2)$ we have
$$f'(a)ge ba^{by-1}(xy^{b(x-y)}-y)=ba^{by-1}y^{b(x-y)}(x-y^{1-b(x-y)})ge 0
$$
so
$$f(x)ge f(y)Longrightarrow x^{bx}+y^{by}ge x^{by}+y^{bx}$$
let $b=1$
we have
$$x^x+y^yge x^y+y^x$$
answered Jun 4 '13 at 12:21
math110
32.4k455215
The differentiation step does not look right to me, we have either $ (a^x)' = ln (a) times a^x $ or $(x^a)'=a times x^{a-1}$.
– Viet Hoang Quoc
Jun 5 '13 at 8:15
I use $(x^a)'_{x}=atimes x^{a-1}$..
– math110
Sep 3 '13 at 0:43
add a comment |
up vote
1
down vote
without loss of generality we only prove by $0le yle xle 1$,let
$$f(a)=a^{bx}-a^{by},xge age y,1ge bx-byge 0$$
then
$$(a^{bx-by})'_{a}=dfrac{bx-by}{a}cdot a^{bx-by}>0,Longrightarrow a^{bx-by}ge y^{bx-by}cdots (1)$$
since use $AM-GM$ inequality,we have
$$y^{1+y-x}1^{x+xy-y^2-y}leleft(dfrac{x}{1+xy-y^2}right)^{1+xy-y^2}le xcdots (2)$$
and
$$f'(a)=dfrac{bx}{a}cdot a^{bx}-dfrac{by}{a}cdot a^{by}=dfrac{ba^{by}}{a}(xa^{b(x-y)}-y)=$$
so use $(1),(2)$ we have
$$f'(a)ge ba^{by-1}(xy^{b(x-y)}-y)=ba^{by-1}y^{b(x-y)}(x-y^{1-b(x-y)})ge 0
$$
so
$$f(x)ge f(y)Longrightarrow x^{bx}+y^{by}ge x^{by}+y^{bx}$$
let $b=1$
we have
$$x^x+y^yge x^y+y^x$$
answered Jun 4 '13 at 12:21
math110
32.4k455215
The differentiation step does not look right to me, we have either $ (a^x)' = ln (a) times a^x $ or $(x^a)'=a times x^{a-1}$.
– Viet Hoang Quoc
Jun 5 '13 at 8:15
I use $(x^a)'_{x}=atimes x^{a-1}$..
– math110
Sep 3 '13 at 0:43
add a comment |
up vote
1
down vote
up vote
1
down vote
without loss of generality we only prove by $0le yle xle 1$,let
$$f(a)=a^{bx}-a^{by},xge age y,1ge bx-byge 0$$
then
$$(a^{bx-by})'_{a}=dfrac{bx-by}{a}cdot a^{bx-by}>0,Longrightarrow a^{bx-by}ge y^{bx-by}cdots (1)$$
since use $AM-GM$ inequality,we have
$$y^{1+y-x}1^{x+xy-y^2-y}leleft(dfrac{x}{1+xy-y^2}right)^{1+xy-y^2}le xcdots (2)$$
and
$$f'(a)=dfrac{bx}{a}cdot a^{bx}-dfrac{by}{a}cdot a^{by}=dfrac{ba^{by}}{a}(xa^{b(x-y)}-y)=$$
so use $(1),(2)$ we have
$$f'(a)ge ba^{by-1}(xy^{b(x-y)}-y)=ba^{by-1}y^{b(x-y)}(x-y^{1-b(x-y)})ge 0
$$
so
$$f(x)ge f(y)Longrightarrow x^{bx}+y^{by}ge x^{by}+y^{bx}$$
let $b=1$
we have
$$x^x+y^yge x^y+y^x$$
answered Jun 4 '13 at 12:21
math110
32.4k455215
without loss of generality we only prove by $0le yle xle 1$,let
$$f(a)=a^{bx}-a^{by},xge age y,1ge bx-byge 0$$
then
$$(a^{bx-by})'_{a}=dfrac{bx-by}{a}cdot a^{bx-by}>0,Longrightarrow a^{bx-by}ge y^{bx-by}cdots (1)$$
since use $AM-GM$ inequality,we have
$$y^{1+y-x}1^{x+xy-y^2-y}leleft(dfrac{x}{1+xy-y^2}right)^{1+xy-y^2}le xcdots (2)$$
and
$$f'(a)=dfrac{bx}{a}cdot a^{bx}-dfrac{by}{a}cdot a^{by}=dfrac{ba^{by}}{a}(xa^{b(x-y)}-y)=$$
so use $(1),(2)$ we have
$$f'(a)ge ba^{by-1}(xy^{b(x-y)}-y)=ba^{by-1}y^{b(x-y)}(x-y^{1-b(x-y)})ge 0
$$
so
$$f(x)ge f(y)Longrightarrow x^{bx}+y^{by}ge x^{by}+y^{bx}$$
let $b=1$
we have
$$x^x+y^yge x^y+y^x$$
answered Jun 4 '13 at 12:21
math110
32.4k455215
answered Jun 4 '13 at 12:21
math110
32.4k455215
answered Jun 4 '13 at 12:21
math110
32.4k455215
answered Jun 4 '13 at 12:21
math110
32.4k455215
32.4k455215
The differentiation step does not look right to me, we have either $ (a^x)' = ln (a) times a^x $ or $(x^a)'=a times x^{a-1}$.
– Viet Hoang Quoc
Jun 5 '13 at 8:15
I use $(x^a)'_{x}=atimes x^{a-1}$..
– math110
Sep 3 '13 at 0:43
add a comment |
The differentiation step does not look right to me, we have either $ (a^x)' = ln (a) times a^x $ or $(x^a)'=a times x^{a-1}$.
– Viet Hoang Quoc
Jun 5 '13 at 8:15
I use $(x^a)'_{x}=atimes x^{a-1}$..
– math110
Sep 3 '13 at 0:43
The differentiation step does not look right to me, we have either $ (a^x)' = ln (a) times a^x $ or $(x^a)'=a times x^{a-1}$.
– Viet Hoang Quoc
Jun 5 '13 at 8:15
The differentiation step does not look right to me, we have either $ (a^x)' = ln (a) times a^x $ or $(x^a)'=a times x^{a-1}$.
– Viet Hoang Quoc
Jun 5 '13 at 8:15
I use $(x^a)'_{x}=atimes x^{a-1}$..
– math110
Sep 3 '13 at 0:43
I use $(x^a)'_{x}=atimes x^{a-1}$..
– math110
Sep 3 '13 at 0:43
add a comment |
up vote
1
down vote
Let $xgeq y$.
We'll consider two cases.
$xgeq1$.
Hence, $$x^x+y^y=x^yleft(x^{x-y}-1right)+x^y+y^ygeq y^yleft(x^{x-y}-1right)+x^y+y^ygeq$$
$$geq y^yleft(y^{x-y}-1right)+x^y+y^ygeq x^y+y^x.$$
2. $1>xgeq y>0.$
Let $f(t)=t^x-t^y$, where $tin[y,1).$
Thus, by Bernoulli we obtain:
$$f'(t)=xt^{x-1}-yt^{y-1}=t^{y-1}left(xt^{x-y}-yright)geq$$
$$geq t^{y-1}left(xy^{x-y}-yright)=t^{y-1}y^{x-y}left(x-y^{1-x+y}right)=$$
$$=t^{y-1}y^{x-y}left(x-(1+(y-1))^{1-x+y}right)geq$$
$$geq t^{y-1}y^{x-y}left(x-(1+(y-1)(1-x+y))right)=t^{y-1}y^{x-y}y(x-y)geq0,$$
which gives $$f(x)geq f(y)$$ or
$$x^x-x^ygeq y^x-y^y$$ or
$$x^x+y^ygeq x^y+y^x$$ and we are done!
answered 2 days ago
Michael Rozenberg
94.2k1588183
add a comment |
up vote
1
down vote
Let $xgeq y$.
We'll consider two cases.
$xgeq1$.
Hence, $$x^x+y^y=x^yleft(x^{x-y}-1right)+x^y+y^ygeq y^yleft(x^{x-y}-1right)+x^y+y^ygeq$$
$$geq y^yleft(y^{x-y}-1right)+x^y+y^ygeq x^y+y^x.$$
2. $1>xgeq y>0.$
Let $f(t)=t^x-t^y$, where $tin[y,1).$
Thus, by Bernoulli we obtain:
$$f'(t)=xt^{x-1}-yt^{y-1}=t^{y-1}left(xt^{x-y}-yright)geq$$
$$geq t^{y-1}left(xy^{x-y}-yright)=t^{y-1}y^{x-y}left(x-y^{1-x+y}right)=$$
$$=t^{y-1}y^{x-y}left(x-(1+(y-1))^{1-x+y}right)geq$$
$$geq t^{y-1}y^{x-y}left(x-(1+(y-1)(1-x+y))right)=t^{y-1}y^{x-y}y(x-y)geq0,$$
which gives $$f(x)geq f(y)$$ or
$$x^x-x^ygeq y^x-y^y$$ or
$$x^x+y^ygeq x^y+y^x$$ and we are done!
answered 2 days ago
Michael Rozenberg
94.2k1588183
add a comment |
up vote
1
down vote
up vote
1
down vote
Let $xgeq y$.
We'll consider two cases.
$xgeq1$.
Hence, $$x^x+y^y=x^yleft(x^{x-y}-1right)+x^y+y^ygeq y^yleft(x^{x-y}-1right)+x^y+y^ygeq$$
$$geq y^yleft(y^{x-y}-1right)+x^y+y^ygeq x^y+y^x.$$
2. $1>xgeq y>0.$
Let $f(t)=t^x-t^y$, where $tin[y,1).$
Thus, by Bernoulli we obtain:
$$f'(t)=xt^{x-1}-yt^{y-1}=t^{y-1}left(xt^{x-y}-yright)geq$$
$$geq t^{y-1}left(xy^{x-y}-yright)=t^{y-1}y^{x-y}left(x-y^{1-x+y}right)=$$
$$=t^{y-1}y^{x-y}left(x-(1+(y-1))^{1-x+y}right)geq$$
$$geq t^{y-1}y^{x-y}left(x-(1+(y-1)(1-x+y))right)=t^{y-1}y^{x-y}y(x-y)geq0,$$
which gives $$f(x)geq f(y)$$ or
$$x^x-x^ygeq y^x-y^y$$ or
$$x^x+y^ygeq x^y+y^x$$ and we are done!
answered 2 days ago
Michael Rozenberg
94.2k1588183
Let $xgeq y$.
We'll consider two cases.
$xgeq1$.
Hence, $$x^x+y^y=x^yleft(x^{x-y}-1right)+x^y+y^ygeq y^yleft(x^{x-y}-1right)+x^y+y^ygeq$$
$$geq y^yleft(y^{x-y}-1right)+x^y+y^ygeq x^y+y^x.$$
2. $1>xgeq y>0.$
Let $f(t)=t^x-t^y$, where $tin[y,1).$
Thus, by Bernoulli we obtain:
$$f'(t)=xt^{x-1}-yt^{y-1}=t^{y-1}left(xt^{x-y}-yright)geq$$
$$geq t^{y-1}left(xy^{x-y}-yright)=t^{y-1}y^{x-y}left(x-y^{1-x+y}right)=$$
$$=t^{y-1}y^{x-y}left(x-(1+(y-1))^{1-x+y}right)geq$$
$$geq t^{y-1}y^{x-y}left(x-(1+(y-1)(1-x+y))right)=t^{y-1}y^{x-y}y(x-y)geq0,$$
which gives $$f(x)geq f(y)$$ or
$$x^x-x^ygeq y^x-y^y$$ or
$$x^x+y^ygeq x^y+y^x$$ and we are done!
answered 2 days ago
Michael Rozenberg
94.2k1588183
answered 2 days ago
Michael Rozenberg
94.2k1588183
answered 2 days ago
Michael Rozenberg
94.2k1588183
answered 2 days ago
Michael Rozenberg
94.2k1588183
94.2k1588183
add a comment |
add a comment |
up vote
0
down vote
Assume $xge y$ by symmetry. We want
$$x^x-x^yge y^x-y^y.$$
i.e. $$x^y(x^{x-y}-1)ge y^y(y^{x-y}-1),$$
which is then easy to show by listing all possible cases according to how $x,y$ compare to 1.
answered Jun 4 '13 at 11:38
Spook
2,2121033
add a comment |
up vote
0
down vote
Assume $xge y$ by symmetry. We want
$$x^x-x^yge y^x-y^y.$$
i.e. $$x^y(x^{x-y}-1)ge y^y(y^{x-y}-1),$$
which is then easy to show by listing all possible cases according to how $x,y$ compare to 1.
answered Jun 4 '13 at 11:38
Spook
2,2121033
add a comment |
up vote
0
down vote
up vote
0
down vote
Assume $xge y$ by symmetry. We want
$$x^x-x^yge y^x-y^y.$$
i.e. $$x^y(x^{x-y}-1)ge y^y(y^{x-y}-1),$$
which is then easy to show by listing all possible cases according to how $x,y$ compare to 1.
answered Jun 4 '13 at 11:38
Spook
2,2121033
Assume $xge y$ by symmetry. We want
$$x^x-x^yge y^x-y^y.$$
i.e. $$x^y(x^{x-y}-1)ge y^y(y^{x-y}-1),$$
which is then easy to show by listing all possible cases according to how $x,y$ compare to 1.
answered Jun 4 '13 at 11:38
Spook
2,2121033
answered Jun 4 '13 at 11:38
Spook
2,2121033
answered Jun 4 '13 at 11:38
Spook
2,2121033
answered Jun 4 '13 at 11:38
Spook
2,2121033
2,2121033
add a comment |
add a comment |
up vote
-3
down vote
Use induction, taking the base case x=0
answered Jun 4 '13 at 11:21
stancamp1
1
3
This happens for all positive real number and induction clearly does not work nicely here.
– Viet Hoang Quoc
Jun 4 '13 at 11:29
2
This is wrong in the general case. @stancamp1, you may want to edit this answer or delete it in order to avoid downvotes.
– DonAntonio
Jun 4 '13 at 11:39
add a comment |
up vote
-3
down vote
Use induction, taking the base case x=0
answered Jun 4 '13 at 11:21
stancamp1
1
3
This happens for all positive real number and induction clearly does not work nicely here.
– Viet Hoang Quoc
Jun 4 '13 at 11:29
2
This is wrong in the general case. @stancamp1, you may want to edit this answer or delete it in order to avoid downvotes.
– DonAntonio
Jun 4 '13 at 11:39
add a comment |
up vote
-3
down vote
up vote
-3
down vote
Use induction, taking the base case x=0
answered Jun 4 '13 at 11:21
stancamp1
1
Use induction, taking the base case x=0
answered Jun 4 '13 at 11:21
stancamp1
1
answered Jun 4 '13 at 11:21
stancamp1
1
answered Jun 4 '13 at 11:21
stancamp1
1
answered Jun 4 '13 at 11:21
stancamp1
1
1
3
This happens for all positive real number and induction clearly does not work nicely here.
– Viet Hoang Quoc
Jun 4 '13 at 11:29
2
This is wrong in the general case. @stancamp1, you may want to edit this answer or delete it in order to avoid downvotes.
– DonAntonio
Jun 4 '13 at 11:39
add a comment |
3
This happens for all positive real number and induction clearly does not work nicely here.
– Viet Hoang Quoc
Jun 4 '13 at 11:29
2
This is wrong in the general case. @stancamp1, you may want to edit this answer or delete it in order to avoid downvotes.
– DonAntonio
Jun 4 '13 at 11:39
3
3
This happens for all positive real number and induction clearly does not work nicely here.
– Viet Hoang Quoc
Jun 4 '13 at 11:29
This happens for all positive real number and induction clearly does not work nicely here.
– Viet Hoang Quoc
Jun 4 '13 at 11:29
2
2
This is wrong in the general case. @stancamp1, you may want to edit this answer or delete it in order to avoid downvotes.
– DonAntonio
Jun 4 '13 at 11:39
This is wrong in the general case. @stancamp1, you may want to edit this answer or delete it in order to avoid downvotes.
– DonAntonio
Jun 4 '13 at 11:39
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:use $AM-GM$ inequality or define function $(f(x)=x^x)$
– M.H
Jun 4 '13 at 11:03
Can you elaborate a bit more because I have used AM-GM but the difference in the power could be a major issue?
– Viet Hoang Quoc
Jun 4 '13 at 11:13
There is a long ML thread about this and related ones at artofproblemsolving.com/Forum/viewtopic.php?f=52&t=118722
– Wolfgang
Oct 12 '13 at 16:41