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For positive real numbers $x,y,z$, Prove that
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For positive real numbers $x,y,z$, Prove that,
$$bigg(fracx^2+y^2+z^2x+y+zbigg)^(x+y+z)geq x^xy^yz^z geq
bigg(fracx+y+z3bigg)^(x+y+z)$$
I have no idea how to begin this? I am getting a hint for A.M. - G.M. equality but numbers don't seem to fit.
Thanks for you time.
sequences-and-series exponential-function jensen-inequality
edited Aug 1 at 17:20
user 108128
18.8k41544
asked Aug 1 at 16:46
Stacy Barrymore
305
add a comment |Â
up vote
2
down vote
favorite
For positive real numbers $x,y,z$, Prove that,
$$bigg(fracx^2+y^2+z^2x+y+zbigg)^(x+y+z)geq x^xy^yz^z geq
bigg(fracx+y+z3bigg)^(x+y+z)$$
I have no idea how to begin this? I am getting a hint for A.M. - G.M. equality but numbers don't seem to fit.
Thanks for you time.
sequences-and-series exponential-function jensen-inequality
edited Aug 1 at 17:20
user 108128
18.8k41544
asked Aug 1 at 16:46
Stacy Barrymore
305
Stacy: Where did you see this problem ? in a magazine ? a book or a homework. Which class you are taking that the Prof. assigned this problem? it's interesting.
– DeepSea
Aug 1 at 17:11
@DeepSea This was problem in an assignment given by our professor.
– Stacy Barrymore
Aug 1 at 17:15
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
For positive real numbers $x,y,z$, Prove that,
$$bigg(fracx^2+y^2+z^2x+y+zbigg)^(x+y+z)geq x^xy^yz^z geq
bigg(fracx+y+z3bigg)^(x+y+z)$$
I have no idea how to begin this? I am getting a hint for A.M. - G.M. equality but numbers don't seem to fit.
Thanks for you time.
sequences-and-series exponential-function jensen-inequality
edited Aug 1 at 17:20
user 108128
18.8k41544
asked Aug 1 at 16:46
Stacy Barrymore
305
For positive real numbers $x,y,z$, Prove that,
$$bigg(fracx^2+y^2+z^2x+y+zbigg)^(x+y+z)geq x^xy^yz^z geq
bigg(fracx+y+z3bigg)^(x+y+z)$$
I have no idea how to begin this? I am getting a hint for A.M. - G.M. equality but numbers don't seem to fit.
Thanks for you time.
sequences-and-series exponential-function jensen-inequality
edited Aug 1 at 17:20
user 108128
18.8k41544
asked Aug 1 at 16:46
Stacy Barrymore
305
edited Aug 1 at 17:20
user 108128
18.8k41544
edited Aug 1 at 17:20
user 108128
18.8k41544
edited Aug 1 at 17:20
user 108128
18.8k41544
18.8k41544
asked Aug 1 at 16:46
Stacy Barrymore
305
asked Aug 1 at 16:46
Stacy Barrymore
305
asked Aug 1 at 16:46
Stacy Barrymore
305
305
Stacy: Where did you see this problem ? in a magazine ? a book or a homework. Which class you are taking that the Prof. assigned this problem? it's interesting.
– DeepSea
Aug 1 at 17:11
@DeepSea This was problem in an assignment given by our professor.
– Stacy Barrymore
Aug 1 at 17:15
add a comment |Â
Stacy: Where did you see this problem ? in a magazine ? a book or a homework. Which class you are taking that the Prof. assigned this problem? it's interesting.
– DeepSea
Aug 1 at 17:11
@DeepSea This was problem in an assignment given by our professor.
– Stacy Barrymore
Aug 1 at 17:15
Stacy: Where did you see this problem ? in a magazine ? a book or a homework. Which class you are taking that the Prof. assigned this problem? it's interesting.
– DeepSea
Aug 1 at 17:11
Stacy: Where did you see this problem ? in a magazine ? a book or a homework. Which class you are taking that the Prof. assigned this problem? it's interesting.
– DeepSea
Aug 1 at 17:11
@DeepSea This was problem in an assignment given by our professor.
– Stacy Barrymore
Aug 1 at 17:15
@DeepSea This was problem in an assignment given by our professor.
– Stacy Barrymore
Aug 1 at 17:15
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
HINT:
Try using $x$ times $x$, $y$ times $y$ , $z$ times $z$.
EDIT:
Seemingly people downvote hints so updating the full solution.
Using A.M. - G.M. inequality,
$$fracx+x+..(x ;Bbbtimes)+y+y+..(y ;Bbbtimes)+z+z+..(z ;Bbbtimes)x+y+zgeq (x^xy^yz^z)^frac1x+y+Z$$
$$bigg(fracx^2+y^2+z^2x+y+zbigg)^(x+y+z)geq x^xy^yz^z $$
Using G.M. - H.M. inequality,
$$(x^xy^yz^z)^frac1x+y+Z geq fracx+y+zfrac1x+frac1x+..(x ; Bbbtimes)+frac1y+frac1y+..(y ; Bbbtimes)+frac1z+frac1z+..(z ; Bbbtimes)$$
$$ x^xy^yz^z geq
bigg(fracx+y+z3bigg)^(x+y+z)$$
$$bigg(fracx^2+y^2+z^2x+y+zbigg)^(x+y+z)geq x^xy^yz^z geq
bigg(fracx+y+z3bigg)^(x+y+z)$$
answered Aug 1 at 16:47
prog_SAHIL
773217
How did you type the answer that fast? You posted the answer within $2$ minutes the question was posted
– tien lee
Aug 1 at 16:51
@tienlee copied authors notation and I work on a PC so.
– prog_SAHIL
Aug 1 at 16:51
What does mean $pi$ times?
– user 108128
Aug 1 at 16:52
Better to just refer to the Weighted AM-GM Inequality. As @user108128, "$pi$ copies of $pi$" does not really mean much.
– Batominovski
Aug 1 at 16:53
@user108128 just to quantify the number of terms, nothing more.
– prog_SAHIL
Aug 1 at 16:55
 |Â
show 1 more comment
up vote
4
down vote
Since $ln$ is a concave function we obtain:
$$sum_cycfracxx+y+zlnxleqlnleft(sum_cycfracxx+y+zcdot xright)=lnfracx^2+y^2+z^2x+y+z,$$
which gives a left inequality.
The right inequality it's just Jensen for the convex function $f(x)=xlnx:$
$$fracsumlimits_cycxlnx3geqfracx+y+z3lnfracx+y+z3.$$
answered Aug 1 at 16:57
Michael Rozenberg
87.3k1577179
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
HINT:
Try using $x$ times $x$, $y$ times $y$ , $z$ times $z$.
EDIT:
Seemingly people downvote hints so updating the full solution.
Using A.M. - G.M. inequality,
$$fracx+x+..(x ;Bbbtimes)+y+y+..(y ;Bbbtimes)+z+z+..(z ;Bbbtimes)x+y+zgeq (x^xy^yz^z)^frac1x+y+Z$$
$$bigg(fracx^2+y^2+z^2x+y+zbigg)^(x+y+z)geq x^xy^yz^z $$
Using G.M. - H.M. inequality,
$$(x^xy^yz^z)^frac1x+y+Z geq fracx+y+zfrac1x+frac1x+..(x ; Bbbtimes)+frac1y+frac1y+..(y ; Bbbtimes)+frac1z+frac1z+..(z ; Bbbtimes)$$
$$ x^xy^yz^z geq
bigg(fracx+y+z3bigg)^(x+y+z)$$
$$bigg(fracx^2+y^2+z^2x+y+zbigg)^(x+y+z)geq x^xy^yz^z geq
bigg(fracx+y+z3bigg)^(x+y+z)$$
answered Aug 1 at 16:47
prog_SAHIL
773217
How did you type the answer that fast? You posted the answer within $2$ minutes the question was posted
– tien lee
Aug 1 at 16:51
@tienlee copied authors notation and I work on a PC so.
– prog_SAHIL
Aug 1 at 16:51
What does mean $pi$ times?
– user 108128
Aug 1 at 16:52
Better to just refer to the Weighted AM-GM Inequality. As @user108128, "$pi$ copies of $pi$" does not really mean much.
– Batominovski
Aug 1 at 16:53
@user108128 just to quantify the number of terms, nothing more.
– prog_SAHIL
Aug 1 at 16:55
 |Â
show 1 more comment
up vote
2
down vote
accepted
HINT:
Try using $x$ times $x$, $y$ times $y$ , $z$ times $z$.
EDIT:
Seemingly people downvote hints so updating the full solution.
Using A.M. - G.M. inequality,
$$fracx+x+..(x ;Bbbtimes)+y+y+..(y ;Bbbtimes)+z+z+..(z ;Bbbtimes)x+y+zgeq (x^xy^yz^z)^frac1x+y+Z$$
$$bigg(fracx^2+y^2+z^2x+y+zbigg)^(x+y+z)geq x^xy^yz^z $$
Using G.M. - H.M. inequality,
$$(x^xy^yz^z)^frac1x+y+Z geq fracx+y+zfrac1x+frac1x+..(x ; Bbbtimes)+frac1y+frac1y+..(y ; Bbbtimes)+frac1z+frac1z+..(z ; Bbbtimes)$$
$$ x^xy^yz^z geq
bigg(fracx+y+z3bigg)^(x+y+z)$$
$$bigg(fracx^2+y^2+z^2x+y+zbigg)^(x+y+z)geq x^xy^yz^z geq
bigg(fracx+y+z3bigg)^(x+y+z)$$
answered Aug 1 at 16:47
prog_SAHIL
773217
How did you type the answer that fast? You posted the answer within $2$ minutes the question was posted
– tien lee
Aug 1 at 16:51
@tienlee copied authors notation and I work on a PC so.
– prog_SAHIL
Aug 1 at 16:51
What does mean $pi$ times?
– user 108128
Aug 1 at 16:52
Better to just refer to the Weighted AM-GM Inequality. As @user108128, "$pi$ copies of $pi$" does not really mean much.
– Batominovski
Aug 1 at 16:53
@user108128 just to quantify the number of terms, nothing more.
– prog_SAHIL
Aug 1 at 16:55
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
HINT:
Try using $x$ times $x$, $y$ times $y$ , $z$ times $z$.
EDIT:
Seemingly people downvote hints so updating the full solution.
Using A.M. - G.M. inequality,
$$fracx+x+..(x ;Bbbtimes)+y+y+..(y ;Bbbtimes)+z+z+..(z ;Bbbtimes)x+y+zgeq (x^xy^yz^z)^frac1x+y+Z$$
$$bigg(fracx^2+y^2+z^2x+y+zbigg)^(x+y+z)geq x^xy^yz^z $$
Using G.M. - H.M. inequality,
$$(x^xy^yz^z)^frac1x+y+Z geq fracx+y+zfrac1x+frac1x+..(x ; Bbbtimes)+frac1y+frac1y+..(y ; Bbbtimes)+frac1z+frac1z+..(z ; Bbbtimes)$$
$$ x^xy^yz^z geq
bigg(fracx+y+z3bigg)^(x+y+z)$$
$$bigg(fracx^2+y^2+z^2x+y+zbigg)^(x+y+z)geq x^xy^yz^z geq
bigg(fracx+y+z3bigg)^(x+y+z)$$
answered Aug 1 at 16:47
prog_SAHIL
773217
HINT:
Try using $x$ times $x$, $y$ times $y$ , $z$ times $z$.
EDIT:
Seemingly people downvote hints so updating the full solution.
Using A.M. - G.M. inequality,
$$fracx+x+..(x ;Bbbtimes)+y+y+..(y ;Bbbtimes)+z+z+..(z ;Bbbtimes)x+y+zgeq (x^xy^yz^z)^frac1x+y+Z$$
$$bigg(fracx^2+y^2+z^2x+y+zbigg)^(x+y+z)geq x^xy^yz^z $$
Using G.M. - H.M. inequality,
$$(x^xy^yz^z)^frac1x+y+Z geq fracx+y+zfrac1x+frac1x+..(x ; Bbbtimes)+frac1y+frac1y+..(y ; Bbbtimes)+frac1z+frac1z+..(z ; Bbbtimes)$$
$$ x^xy^yz^z geq
bigg(fracx+y+z3bigg)^(x+y+z)$$
$$bigg(fracx^2+y^2+z^2x+y+zbigg)^(x+y+z)geq x^xy^yz^z geq
bigg(fracx+y+z3bigg)^(x+y+z)$$
answered Aug 1 at 16:47
prog_SAHIL
773217
answered Aug 1 at 16:47
prog_SAHIL
773217
answered Aug 1 at 16:47
prog_SAHIL
773217
answered Aug 1 at 16:47
prog_SAHIL
773217
773217
How did you type the answer that fast? You posted the answer within $2$ minutes the question was posted
– tien lee
Aug 1 at 16:51
@tienlee copied authors notation and I work on a PC so.
– prog_SAHIL
Aug 1 at 16:51
What does mean $pi$ times?
– user 108128
Aug 1 at 16:52
Better to just refer to the Weighted AM-GM Inequality. As @user108128, "$pi$ copies of $pi$" does not really mean much.
– Batominovski
Aug 1 at 16:53
@user108128 just to quantify the number of terms, nothing more.
– prog_SAHIL
Aug 1 at 16:55
 |Â
show 1 more comment
How did you type the answer that fast? You posted the answer within $2$ minutes the question was posted
– tien lee
Aug 1 at 16:51
@tienlee copied authors notation and I work on a PC so.
– prog_SAHIL
Aug 1 at 16:51
What does mean $pi$ times?
– user 108128
Aug 1 at 16:52
Better to just refer to the Weighted AM-GM Inequality. As @user108128, "$pi$ copies of $pi$" does not really mean much.
– Batominovski
Aug 1 at 16:53
@user108128 just to quantify the number of terms, nothing more.
– prog_SAHIL
Aug 1 at 16:55
How did you type the answer that fast? You posted the answer within $2$ minutes the question was posted
– tien lee
Aug 1 at 16:51
How did you type the answer that fast? You posted the answer within $2$ minutes the question was posted
– tien lee
Aug 1 at 16:51
@tienlee copied authors notation and I work on a PC so.
– prog_SAHIL
Aug 1 at 16:51
@tienlee copied authors notation and I work on a PC so.
– prog_SAHIL
Aug 1 at 16:51
What does mean $pi$ times?
– user 108128
Aug 1 at 16:52
What does mean $pi$ times?
– user 108128
Aug 1 at 16:52
Better to just refer to the Weighted AM-GM Inequality. As @user108128, "$pi$ copies of $pi$" does not really mean much.
– Batominovski
Aug 1 at 16:53
Better to just refer to the Weighted AM-GM Inequality. As @user108128, "$pi$ copies of $pi$" does not really mean much.
– Batominovski
Aug 1 at 16:53
@user108128 just to quantify the number of terms, nothing more.
– prog_SAHIL
Aug 1 at 16:55
@user108128 just to quantify the number of terms, nothing more.
– prog_SAHIL
Aug 1 at 16:55
 |Â
show 1 more comment
up vote
4
down vote
Since $ln$ is a concave function we obtain:
$$sum_cycfracxx+y+zlnxleqlnleft(sum_cycfracxx+y+zcdot xright)=lnfracx^2+y^2+z^2x+y+z,$$
which gives a left inequality.
The right inequality it's just Jensen for the convex function $f(x)=xlnx:$
$$fracsumlimits_cycxlnx3geqfracx+y+z3lnfracx+y+z3.$$
answered Aug 1 at 16:57
Michael Rozenberg
87.3k1577179
add a comment |Â
up vote
4
down vote
Since $ln$ is a concave function we obtain:
$$sum_cycfracxx+y+zlnxleqlnleft(sum_cycfracxx+y+zcdot xright)=lnfracx^2+y^2+z^2x+y+z,$$
which gives a left inequality.
The right inequality it's just Jensen for the convex function $f(x)=xlnx:$
$$fracsumlimits_cycxlnx3geqfracx+y+z3lnfracx+y+z3.$$
answered Aug 1 at 16:57
Michael Rozenberg
87.3k1577179
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Since $ln$ is a concave function we obtain:
$$sum_cycfracxx+y+zlnxleqlnleft(sum_cycfracxx+y+zcdot xright)=lnfracx^2+y^2+z^2x+y+z,$$
which gives a left inequality.
The right inequality it's just Jensen for the convex function $f(x)=xlnx:$
$$fracsumlimits_cycxlnx3geqfracx+y+z3lnfracx+y+z3.$$
answered Aug 1 at 16:57
Michael Rozenberg
87.3k1577179
Since $ln$ is a concave function we obtain:
$$sum_cycfracxx+y+zlnxleqlnleft(sum_cycfracxx+y+zcdot xright)=lnfracx^2+y^2+z^2x+y+z,$$
which gives a left inequality.
The right inequality it's just Jensen for the convex function $f(x)=xlnx:$
$$fracsumlimits_cycxlnx3geqfracx+y+z3lnfracx+y+z3.$$
answered Aug 1 at 16:57
Michael Rozenberg
87.3k1577179
answered Aug 1 at 16:57
Michael Rozenberg
87.3k1577179
answered Aug 1 at 16:57
Michael Rozenberg
87.3k1577179
answered Aug 1 at 16:57
Michael Rozenberg
87.3k1577179
87.3k1577179
add a comment |Â
add a comment |Â
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Stacy: Where did you see this problem ? in a magazine ? a book or a homework. Which class you are taking that the Prof. assigned this problem? it's interesting.
– DeepSea
Aug 1 at 17:11
@DeepSea This was problem in an assignment given by our professor.
– Stacy Barrymore
Aug 1 at 17:15