Mixing
Showing that if $p_1 + cdots p_n = 1$ then $displaystyle sum_k=1^n left(p_k + dfrac 1p_k right)^2 ge n^3+2n+dfrac 1n$?
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This problem is from the book "Cauchy-Schwarz Masterclass":
Show that if $p_1 + cdots p_n = 1$ with each $p_i$ positive, then $displaystyle sum_k=1^n left(p_k + dfrac 1p_k right)^2 ge n^3+2n+dfrac 1n$
I expanded the LHS and arrived at
$$(p_1^2 + cdots + p_n^2) + left(dfrac 1p_1^2 + cdots + dfrac 1p_n^2right) ge n^3 + dfrac 1n$$
I was able to show that $p_1^2 + cdots + p_n^2 ge dfrac 1n$ by applying C-S to the sequences $(p_1, ..., p_n)$ and $(p_1, ..., p_n)$. I also think that $dfrac 1p_1^2 + cdots + dfrac 1p_n^2 ge n^3$ is true (because equality holds for $p_i = dfrac 1n$ and I checked numerically for some values) but I am not able to prove it.
At this point in the book we only proed that classic $C-S$ inequality and the $C-S$ for inner product space, so I am hoping there is a solution only with these tools.
real-analysis inequality a.m.-g.m.-inequality cauchy-schwarz-inequality
asked Jul 18 at 21:57
Ovi
11.3k935105
add a comment |Â
up vote
5
down vote
favorite
This problem is from the book "Cauchy-Schwarz Masterclass":
Show that if $p_1 + cdots p_n = 1$ with each $p_i$ positive, then $displaystyle sum_k=1^n left(p_k + dfrac 1p_k right)^2 ge n^3+2n+dfrac 1n$
I expanded the LHS and arrived at
$$(p_1^2 + cdots + p_n^2) + left(dfrac 1p_1^2 + cdots + dfrac 1p_n^2right) ge n^3 + dfrac 1n$$
I was able to show that $p_1^2 + cdots + p_n^2 ge dfrac 1n$ by applying C-S to the sequences $(p_1, ..., p_n)$ and $(p_1, ..., p_n)$. I also think that $dfrac 1p_1^2 + cdots + dfrac 1p_n^2 ge n^3$ is true (because equality holds for $p_i = dfrac 1n$ and I checked numerically for some values) but I am not able to prove it.
At this point in the book we only proed that classic $C-S$ inequality and the $C-S$ for inner product space, so I am hoping there is a solution only with these tools.
real-analysis inequality a.m.-g.m.-inequality cauchy-schwarz-inequality
asked Jul 18 at 21:57
Ovi
11.3k935105
1
Did you write the LHS correctly? I'm not sure where you are getting the squares from in your expanded LHS. EDIT: The question was fixed.
– Joe
Jul 18 at 22:00
Do you mean $sum_k=1^n left(p_k + dfrac 1p_k right)^2$? Otherwise this is false - take $p_k = 1/n$.
– marty cohen
Jul 18 at 22:02
A solution is already here: Proving inequality $(a_1+frac1a_1)^2+(a_2+frac1a_2)^2+cdots+(a_n+frac1a_n)^2 ge frac(n^2+1)^2n$ This version of the question is better, IMO.
– mechanodroid
Jul 18 at 22:27
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
This problem is from the book "Cauchy-Schwarz Masterclass":
Show that if $p_1 + cdots p_n = 1$ with each $p_i$ positive, then $displaystyle sum_k=1^n left(p_k + dfrac 1p_k right)^2 ge n^3+2n+dfrac 1n$
I expanded the LHS and arrived at
$$(p_1^2 + cdots + p_n^2) + left(dfrac 1p_1^2 + cdots + dfrac 1p_n^2right) ge n^3 + dfrac 1n$$
I was able to show that $p_1^2 + cdots + p_n^2 ge dfrac 1n$ by applying C-S to the sequences $(p_1, ..., p_n)$ and $(p_1, ..., p_n)$. I also think that $dfrac 1p_1^2 + cdots + dfrac 1p_n^2 ge n^3$ is true (because equality holds for $p_i = dfrac 1n$ and I checked numerically for some values) but I am not able to prove it.
At this point in the book we only proed that classic $C-S$ inequality and the $C-S$ for inner product space, so I am hoping there is a solution only with these tools.
real-analysis inequality a.m.-g.m.-inequality cauchy-schwarz-inequality
asked Jul 18 at 21:57
Ovi
11.3k935105
This problem is from the book "Cauchy-Schwarz Masterclass":
Show that if $p_1 + cdots p_n = 1$ with each $p_i$ positive, then $displaystyle sum_k=1^n left(p_k + dfrac 1p_k right)^2 ge n^3+2n+dfrac 1n$
I expanded the LHS and arrived at
$$(p_1^2 + cdots + p_n^2) + left(dfrac 1p_1^2 + cdots + dfrac 1p_n^2right) ge n^3 + dfrac 1n$$
I was able to show that $p_1^2 + cdots + p_n^2 ge dfrac 1n$ by applying C-S to the sequences $(p_1, ..., p_n)$ and $(p_1, ..., p_n)$. I also think that $dfrac 1p_1^2 + cdots + dfrac 1p_n^2 ge n^3$ is true (because equality holds for $p_i = dfrac 1n$ and I checked numerically for some values) but I am not able to prove it.
At this point in the book we only proed that classic $C-S$ inequality and the $C-S$ for inner product space, so I am hoping there is a solution only with these tools.
real-analysis inequality a.m.-g.m.-inequality cauchy-schwarz-inequality
asked Jul 18 at 21:57
Ovi
11.3k935105
edited Jul 19 at 0:28
asked Jul 18 at 21:57
Ovi
11.3k935105
asked Jul 18 at 21:57
Ovi
11.3k935105
asked Jul 18 at 21:57
Ovi
11.3k935105
11.3k935105
1
Did you write the LHS correctly? I'm not sure where you are getting the squares from in your expanded LHS. EDIT: The question was fixed.
– Joe
Jul 18 at 22:00
Do you mean $sum_k=1^n left(p_k + dfrac 1p_k right)^2$? Otherwise this is false - take $p_k = 1/n$.
– marty cohen
Jul 18 at 22:02
A solution is already here: Proving inequality $(a_1+frac1a_1)^2+(a_2+frac1a_2)^2+cdots+(a_n+frac1a_n)^2 ge frac(n^2+1)^2n$ This version of the question is better, IMO.
– mechanodroid
Jul 18 at 22:27
add a comment |Â
1
Did you write the LHS correctly? I'm not sure where you are getting the squares from in your expanded LHS. EDIT: The question was fixed.
– Joe
Jul 18 at 22:00
Do you mean $sum_k=1^n left(p_k + dfrac 1p_k right)^2$? Otherwise this is false - take $p_k = 1/n$.
– marty cohen
Jul 18 at 22:02
A solution is already here: Proving inequality $(a_1+frac1a_1)^2+(a_2+frac1a_2)^2+cdots+(a_n+frac1a_n)^2 ge frac(n^2+1)^2n$ This version of the question is better, IMO.
– mechanodroid
Jul 18 at 22:27
1
1
Did you write the LHS correctly? I'm not sure where you are getting the squares from in your expanded LHS. EDIT: The question was fixed.
– Joe
Jul 18 at 22:00
Did you write the LHS correctly? I'm not sure where you are getting the squares from in your expanded LHS. EDIT: The question was fixed.
– Joe
Jul 18 at 22:00
Do you mean $sum_k=1^n left(p_k + dfrac 1p_k right)^2$? Otherwise this is false - take $p_k = 1/n$.
– marty cohen
Jul 18 at 22:02
Do you mean $sum_k=1^n left(p_k + dfrac 1p_k right)^2$? Otherwise this is false - take $p_k = 1/n$.
– marty cohen
Jul 18 at 22:02
A solution is already here: Proving inequality $(a_1+frac1a_1)^2+(a_2+frac1a_2)^2+cdots+(a_n+frac1a_n)^2 ge frac(n^2+1)^2n$ This version of the question is better, IMO.
– mechanodroid
Jul 18 at 22:27
A solution is already here: Proving inequality $(a_1+frac1a_1)^2+(a_2+frac1a_2)^2+cdots+(a_n+frac1a_n)^2 ge frac(n^2+1)^2n$ This version of the question is better, IMO.
– mechanodroid
Jul 18 at 22:27
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
10
down vote
accepted
This is a proof that only uses Cauchy Schwarz, the second part can be simplified if you are allowed to use AM $ge$ HM.
Apply CS to $n$ copies of $1$ and $displaystyle;p_k+frac1p_k$, we obtain
$$n sum_k=1^n left(p_k + frac1p_kright)^2
= left( sum_k=1^n 1^2right)sum_k=1^nleft(p_k + frac1p_kright)^2
ge left(sum_k=1^n p_k + frac1p_kright)^2
= left(1 + sum_k=1^n frac1p_kright)^2$$
Apply CS again to $sqrtp_k$ and $displaystyle;frac1sqrtp_k$, we obtain
$$sum_k=1^n frac1p_k = sum_k=1^n sqrtp_k^2 sum_k=1^n frac1sqrtp_k^2
ge left(sum_k=1^n fracsqrtp_ksqrtp_kright)^2 = n^2$$
Combine these two inequalities, we obtain
$$sum_k=1^n left(p_k + frac1p_kright)^2
ge frac1n left(1 + n^2right)^2 = n^3 + 2n + frac1n$$
answered Jul 18 at 22:20
achille hui
91k5127246
+1 thanks for the answer. I wish I got this :( I tried similar things but I didn't get to the solution unfortunately.
– Ovi
Jul 18 at 23:30
add a comment |Â
up vote
4
down vote
Assume all $p_i$ are positive, otherwise the LHS is infinite and the inequality is trivial. Consider the function $fcolon(0,1)tomathbbR$ defined by
$$
f(x) = left(x+frac1xright)^2,qquad xin(0,1) tag1
$$
which is easily seen to be concave. (For instance, $f''(x) = 2+6/x^4 > 0$.)
By Jensen's inequality,
$$
sum_i=1^n frac1n f(p_i) geq fleft(sum_i=1^n fracp_inright) = fleft(frac1nright) tag2
$$
i.e.,
$$
frac1n sum_i=1^n left(p_i+frac1p_iright)^2 geq frac1n^2+2 + n^2,. tag3
$$
Multiply both sides by $n$ to obtain the desired inequality.
answered Jul 18 at 22:17
Clement C.
47.2k33682
add a comment |Â
up vote
2
down vote
In fact we seek to minimize $$dfrac1p_1^2+dfrac1p_2^2+cdots+dfrac1p_n^2$$respect to $$p_1+p_2+cdots+p_n=1$$using Lagrange multipliers we get$$-dfrac2p_1^3=lambda\-dfrac2p_2^3=lambda\.\.\.\-dfrac2p_n^3=lambda$$which yields to $$p_1=p_2=cdots=p_n=dfrac1n$$ and leads to the same result by substitution
answered Jul 18 at 22:03
Mostafa Ayaz
8,6023630
1
And what about the $p_1^2+ldots+p_n^2$?
– amsmath
Jul 18 at 22:45
Also in a similar manner
– Mostafa Ayaz
Jul 19 at 7:19
add a comment |Â
up vote
1
down vote
The Lagrangian reads
$$
L(p,lambda) =sum_k=1^nleft(p_k+frac1p_kright)^2+lambdaleft(sum_k=1^n p_k - 1right)
$$
The stationary onditions are
$$
2p_k-frac2p_k^3+lambda = 0, ;; forall k\
sum_k=1^n p_k = 1
$$
so we conclude $p_1=p_2=cdots= p_n = frac 1n$ hence
$$
sum_k=1^nleft(p_k+frac1p_kright)^2 = nleft(n+frac 1nright)^2 = n^3+2n+frac 1n
$$
answered Jul 19 at 9:44
Cesareo
5,7722412
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
This is a proof that only uses Cauchy Schwarz, the second part can be simplified if you are allowed to use AM $ge$ HM.
Apply CS to $n$ copies of $1$ and $displaystyle;p_k+frac1p_k$, we obtain
$$n sum_k=1^n left(p_k + frac1p_kright)^2
= left( sum_k=1^n 1^2right)sum_k=1^nleft(p_k + frac1p_kright)^2
ge left(sum_k=1^n p_k + frac1p_kright)^2
= left(1 + sum_k=1^n frac1p_kright)^2$$
Apply CS again to $sqrtp_k$ and $displaystyle;frac1sqrtp_k$, we obtain
$$sum_k=1^n frac1p_k = sum_k=1^n sqrtp_k^2 sum_k=1^n frac1sqrtp_k^2
ge left(sum_k=1^n fracsqrtp_ksqrtp_kright)^2 = n^2$$
Combine these two inequalities, we obtain
$$sum_k=1^n left(p_k + frac1p_kright)^2
ge frac1n left(1 + n^2right)^2 = n^3 + 2n + frac1n$$
answered Jul 18 at 22:20
achille hui
91k5127246
+1 thanks for the answer. I wish I got this :( I tried similar things but I didn't get to the solution unfortunately.
– Ovi
Jul 18 at 23:30
add a comment |Â
up vote
10
down vote
accepted
This is a proof that only uses Cauchy Schwarz, the second part can be simplified if you are allowed to use AM $ge$ HM.
Apply CS to $n$ copies of $1$ and $displaystyle;p_k+frac1p_k$, we obtain
$$n sum_k=1^n left(p_k + frac1p_kright)^2
= left( sum_k=1^n 1^2right)sum_k=1^nleft(p_k + frac1p_kright)^2
ge left(sum_k=1^n p_k + frac1p_kright)^2
= left(1 + sum_k=1^n frac1p_kright)^2$$
Apply CS again to $sqrtp_k$ and $displaystyle;frac1sqrtp_k$, we obtain
$$sum_k=1^n frac1p_k = sum_k=1^n sqrtp_k^2 sum_k=1^n frac1sqrtp_k^2
ge left(sum_k=1^n fracsqrtp_ksqrtp_kright)^2 = n^2$$
Combine these two inequalities, we obtain
$$sum_k=1^n left(p_k + frac1p_kright)^2
ge frac1n left(1 + n^2right)^2 = n^3 + 2n + frac1n$$
answered Jul 18 at 22:20
achille hui
91k5127246
+1 thanks for the answer. I wish I got this :( I tried similar things but I didn't get to the solution unfortunately.
– Ovi
Jul 18 at 23:30
add a comment |Â
up vote
10
down vote
accepted
up vote
10
down vote
accepted
This is a proof that only uses Cauchy Schwarz, the second part can be simplified if you are allowed to use AM $ge$ HM.
Apply CS to $n$ copies of $1$ and $displaystyle;p_k+frac1p_k$, we obtain
$$n sum_k=1^n left(p_k + frac1p_kright)^2
= left( sum_k=1^n 1^2right)sum_k=1^nleft(p_k + frac1p_kright)^2
ge left(sum_k=1^n p_k + frac1p_kright)^2
= left(1 + sum_k=1^n frac1p_kright)^2$$
Apply CS again to $sqrtp_k$ and $displaystyle;frac1sqrtp_k$, we obtain
$$sum_k=1^n frac1p_k = sum_k=1^n sqrtp_k^2 sum_k=1^n frac1sqrtp_k^2
ge left(sum_k=1^n fracsqrtp_ksqrtp_kright)^2 = n^2$$
Combine these two inequalities, we obtain
$$sum_k=1^n left(p_k + frac1p_kright)^2
ge frac1n left(1 + n^2right)^2 = n^3 + 2n + frac1n$$
answered Jul 18 at 22:20
achille hui
91k5127246
This is a proof that only uses Cauchy Schwarz, the second part can be simplified if you are allowed to use AM $ge$ HM.
Apply CS to $n$ copies of $1$ and $displaystyle;p_k+frac1p_k$, we obtain
$$n sum_k=1^n left(p_k + frac1p_kright)^2
= left( sum_k=1^n 1^2right)sum_k=1^nleft(p_k + frac1p_kright)^2
ge left(sum_k=1^n p_k + frac1p_kright)^2
= left(1 + sum_k=1^n frac1p_kright)^2$$
Apply CS again to $sqrtp_k$ and $displaystyle;frac1sqrtp_k$, we obtain
$$sum_k=1^n frac1p_k = sum_k=1^n sqrtp_k^2 sum_k=1^n frac1sqrtp_k^2
ge left(sum_k=1^n fracsqrtp_ksqrtp_kright)^2 = n^2$$
Combine these two inequalities, we obtain
$$sum_k=1^n left(p_k + frac1p_kright)^2
ge frac1n left(1 + n^2right)^2 = n^3 + 2n + frac1n$$
answered Jul 18 at 22:20
achille hui
91k5127246
answered Jul 18 at 22:20
achille hui
91k5127246
answered Jul 18 at 22:20
achille hui
91k5127246
answered Jul 18 at 22:20
achille hui
91k5127246
91k5127246
+1 thanks for the answer. I wish I got this :( I tried similar things but I didn't get to the solution unfortunately.
– Ovi
Jul 18 at 23:30
add a comment |Â
+1 thanks for the answer. I wish I got this :( I tried similar things but I didn't get to the solution unfortunately.
– Ovi
Jul 18 at 23:30
+1 thanks for the answer. I wish I got this :( I tried similar things but I didn't get to the solution unfortunately.
– Ovi
Jul 18 at 23:30
+1 thanks for the answer. I wish I got this :( I tried similar things but I didn't get to the solution unfortunately.
– Ovi
Jul 18 at 23:30
add a comment |Â
up vote
4
down vote
Assume all $p_i$ are positive, otherwise the LHS is infinite and the inequality is trivial. Consider the function $fcolon(0,1)tomathbbR$ defined by
$$
f(x) = left(x+frac1xright)^2,qquad xin(0,1) tag1
$$
which is easily seen to be concave. (For instance, $f''(x) = 2+6/x^4 > 0$.)
By Jensen's inequality,
$$
sum_i=1^n frac1n f(p_i) geq fleft(sum_i=1^n fracp_inright) = fleft(frac1nright) tag2
$$
i.e.,
$$
frac1n sum_i=1^n left(p_i+frac1p_iright)^2 geq frac1n^2+2 + n^2,. tag3
$$
Multiply both sides by $n$ to obtain the desired inequality.
answered Jul 18 at 22:17
Clement C.
47.2k33682
add a comment |Â
up vote
4
down vote
Assume all $p_i$ are positive, otherwise the LHS is infinite and the inequality is trivial. Consider the function $fcolon(0,1)tomathbbR$ defined by
$$
f(x) = left(x+frac1xright)^2,qquad xin(0,1) tag1
$$
which is easily seen to be concave. (For instance, $f''(x) = 2+6/x^4 > 0$.)
By Jensen's inequality,
$$
sum_i=1^n frac1n f(p_i) geq fleft(sum_i=1^n fracp_inright) = fleft(frac1nright) tag2
$$
i.e.,
$$
frac1n sum_i=1^n left(p_i+frac1p_iright)^2 geq frac1n^2+2 + n^2,. tag3
$$
Multiply both sides by $n$ to obtain the desired inequality.
answered Jul 18 at 22:17
Clement C.
47.2k33682
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Assume all $p_i$ are positive, otherwise the LHS is infinite and the inequality is trivial. Consider the function $fcolon(0,1)tomathbbR$ defined by
$$
f(x) = left(x+frac1xright)^2,qquad xin(0,1) tag1
$$
which is easily seen to be concave. (For instance, $f''(x) = 2+6/x^4 > 0$.)
By Jensen's inequality,
$$
sum_i=1^n frac1n f(p_i) geq fleft(sum_i=1^n fracp_inright) = fleft(frac1nright) tag2
$$
i.e.,
$$
frac1n sum_i=1^n left(p_i+frac1p_iright)^2 geq frac1n^2+2 + n^2,. tag3
$$
Multiply both sides by $n$ to obtain the desired inequality.
answered Jul 18 at 22:17
Clement C.
47.2k33682
Assume all $p_i$ are positive, otherwise the LHS is infinite and the inequality is trivial. Consider the function $fcolon(0,1)tomathbbR$ defined by
$$
f(x) = left(x+frac1xright)^2,qquad xin(0,1) tag1
$$
which is easily seen to be concave. (For instance, $f''(x) = 2+6/x^4 > 0$.)
By Jensen's inequality,
$$
sum_i=1^n frac1n f(p_i) geq fleft(sum_i=1^n fracp_inright) = fleft(frac1nright) tag2
$$
i.e.,
$$
frac1n sum_i=1^n left(p_i+frac1p_iright)^2 geq frac1n^2+2 + n^2,. tag3
$$
Multiply both sides by $n$ to obtain the desired inequality.
answered Jul 18 at 22:17
Clement C.
47.2k33682
answered Jul 18 at 22:17
Clement C.
47.2k33682
answered Jul 18 at 22:17
Clement C.
47.2k33682
answered Jul 18 at 22:17
Clement C.
47.2k33682
47.2k33682
add a comment |Â
add a comment |Â
up vote
2
down vote
In fact we seek to minimize $$dfrac1p_1^2+dfrac1p_2^2+cdots+dfrac1p_n^2$$respect to $$p_1+p_2+cdots+p_n=1$$using Lagrange multipliers we get$$-dfrac2p_1^3=lambda\-dfrac2p_2^3=lambda\.\.\.\-dfrac2p_n^3=lambda$$which yields to $$p_1=p_2=cdots=p_n=dfrac1n$$ and leads to the same result by substitution
answered Jul 18 at 22:03
Mostafa Ayaz
8,6023630
1
And what about the $p_1^2+ldots+p_n^2$?
– amsmath
Jul 18 at 22:45
Also in a similar manner
– Mostafa Ayaz
Jul 19 at 7:19
add a comment |Â
up vote
2
down vote
In fact we seek to minimize $$dfrac1p_1^2+dfrac1p_2^2+cdots+dfrac1p_n^2$$respect to $$p_1+p_2+cdots+p_n=1$$using Lagrange multipliers we get$$-dfrac2p_1^3=lambda\-dfrac2p_2^3=lambda\.\.\.\-dfrac2p_n^3=lambda$$which yields to $$p_1=p_2=cdots=p_n=dfrac1n$$ and leads to the same result by substitution
answered Jul 18 at 22:03
Mostafa Ayaz
8,6023630
1
And what about the $p_1^2+ldots+p_n^2$?
– amsmath
Jul 18 at 22:45
Also in a similar manner
– Mostafa Ayaz
Jul 19 at 7:19
add a comment |Â
up vote
2
down vote
up vote
2
down vote
In fact we seek to minimize $$dfrac1p_1^2+dfrac1p_2^2+cdots+dfrac1p_n^2$$respect to $$p_1+p_2+cdots+p_n=1$$using Lagrange multipliers we get$$-dfrac2p_1^3=lambda\-dfrac2p_2^3=lambda\.\.\.\-dfrac2p_n^3=lambda$$which yields to $$p_1=p_2=cdots=p_n=dfrac1n$$ and leads to the same result by substitution
answered Jul 18 at 22:03
Mostafa Ayaz
8,6023630
In fact we seek to minimize $$dfrac1p_1^2+dfrac1p_2^2+cdots+dfrac1p_n^2$$respect to $$p_1+p_2+cdots+p_n=1$$using Lagrange multipliers we get$$-dfrac2p_1^3=lambda\-dfrac2p_2^3=lambda\.\.\.\-dfrac2p_n^3=lambda$$which yields to $$p_1=p_2=cdots=p_n=dfrac1n$$ and leads to the same result by substitution
answered Jul 18 at 22:03
Mostafa Ayaz
8,6023630
answered Jul 18 at 22:03
Mostafa Ayaz
8,6023630
answered Jul 18 at 22:03
Mostafa Ayaz
8,6023630
answered Jul 18 at 22:03
Mostafa Ayaz
8,6023630
8,6023630
1
And what about the $p_1^2+ldots+p_n^2$?
– amsmath
Jul 18 at 22:45
Also in a similar manner
– Mostafa Ayaz
Jul 19 at 7:19
add a comment |Â
1
And what about the $p_1^2+ldots+p_n^2$?
– amsmath
Jul 18 at 22:45
Also in a similar manner
– Mostafa Ayaz
Jul 19 at 7:19
1
1
And what about the $p_1^2+ldots+p_n^2$?
– amsmath
Jul 18 at 22:45
And what about the $p_1^2+ldots+p_n^2$?
– amsmath
Jul 18 at 22:45
Also in a similar manner
– Mostafa Ayaz
Jul 19 at 7:19
Also in a similar manner
– Mostafa Ayaz
Jul 19 at 7:19
add a comment |Â
up vote
1
down vote
The Lagrangian reads
$$
L(p,lambda) =sum_k=1^nleft(p_k+frac1p_kright)^2+lambdaleft(sum_k=1^n p_k - 1right)
$$
The stationary onditions are
$$
2p_k-frac2p_k^3+lambda = 0, ;; forall k\
sum_k=1^n p_k = 1
$$
so we conclude $p_1=p_2=cdots= p_n = frac 1n$ hence
$$
sum_k=1^nleft(p_k+frac1p_kright)^2 = nleft(n+frac 1nright)^2 = n^3+2n+frac 1n
$$
answered Jul 19 at 9:44
Cesareo
5,7722412
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up vote
1
down vote
The Lagrangian reads
$$
L(p,lambda) =sum_k=1^nleft(p_k+frac1p_kright)^2+lambdaleft(sum_k=1^n p_k - 1right)
$$
The stationary onditions are
$$
2p_k-frac2p_k^3+lambda = 0, ;; forall k\
sum_k=1^n p_k = 1
$$
so we conclude $p_1=p_2=cdots= p_n = frac 1n$ hence
$$
sum_k=1^nleft(p_k+frac1p_kright)^2 = nleft(n+frac 1nright)^2 = n^3+2n+frac 1n
$$
answered Jul 19 at 9:44
Cesareo
5,7722412
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The Lagrangian reads
$$
L(p,lambda) =sum_k=1^nleft(p_k+frac1p_kright)^2+lambdaleft(sum_k=1^n p_k - 1right)
$$
The stationary onditions are
$$
2p_k-frac2p_k^3+lambda = 0, ;; forall k\
sum_k=1^n p_k = 1
$$
so we conclude $p_1=p_2=cdots= p_n = frac 1n$ hence
$$
sum_k=1^nleft(p_k+frac1p_kright)^2 = nleft(n+frac 1nright)^2 = n^3+2n+frac 1n
$$
answered Jul 19 at 9:44
Cesareo
5,7722412
The Lagrangian reads
$$
L(p,lambda) =sum_k=1^nleft(p_k+frac1p_kright)^2+lambdaleft(sum_k=1^n p_k - 1right)
$$
The stationary onditions are
$$
2p_k-frac2p_k^3+lambda = 0, ;; forall k\
sum_k=1^n p_k = 1
$$
so we conclude $p_1=p_2=cdots= p_n = frac 1n$ hence
$$
sum_k=1^nleft(p_k+frac1p_kright)^2 = nleft(n+frac 1nright)^2 = n^3+2n+frac 1n
$$
answered Jul 19 at 9:44
Cesareo
5,7722412
answered Jul 19 at 9:44
Cesareo
5,7722412
answered Jul 19 at 9:44
Cesareo
5,7722412
answered Jul 19 at 9:44
Cesareo
5,7722412
5,7722412
add a comment |Â
add a comment |Â
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1
Did you write the LHS correctly? I'm not sure where you are getting the squares from in your expanded LHS. EDIT: The question was fixed.
– Joe
Jul 18 at 22:00
Do you mean $sum_k=1^n left(p_k + dfrac 1p_k right)^2$? Otherwise this is false - take $p_k = 1/n$.
– marty cohen
Jul 18 at 22:02
A solution is already here: Proving inequality $(a_1+frac1a_1)^2+(a_2+frac1a_2)^2+cdots+(a_n+frac1a_n)^2 ge frac(n^2+1)^2n$ This version of the question is better, IMO.
– mechanodroid
Jul 18 at 22:27