Prove that $a^{ab}+b^{bc}+c^{cd}+d^{da} geq pi$
17
$begingroup$
If $a,b,c,d >0$, and $a+b+c+d=4$, prove that
$$a^{ab}+b^{bc}+c^{cd}+d^{da} geq pi.$$
I don't think Jensen's inequality will help here, but I think first determining where equality holds will be useful. Maybe taking the logarithm or exponential of both sides will also be useful, but I want to in the end get rid of the plus signs in order to simplify it.
inequality
asked May 6 '16 at 20:49
Puzzled417Puzzled417
3,105527
$endgroup$
|
show 9 more comments
17
$begingroup$
If $a,b,c,d >0$, and $a+b+c+d=4$, prove that
$$a^{ab}+b^{bc}+c^{cd}+d^{da} geq pi.$$
I don't think Jensen's inequality will help here, but I think first determining where equality holds will be useful. Maybe taking the logarithm or exponential of both sides will also be useful, but I want to in the end get rid of the plus signs in order to simplify it.
inequality
asked May 6 '16 at 20:49
Puzzled417Puzzled417
3,105527
$endgroup$
$begingroup$
This reminds me the inequality $a^a+b^b>a^b+b^a$ from Wikipédia. Not sure if it help though...
$endgroup$
– Surb
May 6 '16 at 20:54
$begingroup$
May I ask how you came across this problem? Also, do you know whether the constant $pi$ is optimal?
$endgroup$
– Wojowu
May 6 '16 at 20:55
$begingroup$
@Wojowu It is from a problem solving website. Yes, I believe equality is achieved.
$endgroup$
– Puzzled417
May 6 '16 at 20:59
2
$begingroup$
The best I've got out of Mathematica is3.1605859174508652189…, usingNMinimize[{(a^2)^(a^2 b^2) + (b^2)^(b^2 c^2) + (c^2)^(c^2 d^2) + (d^2)^(d^2 a^2), a != 0 && b != 0 && c != 0 && d != 0 && a^2 + b^2 + c^2 + d^2 == 4}, {a, b, c, d}, WorkingPrecision -> 100]
$endgroup$
– Patrick Stevens
May 6 '16 at 21:26
2
$begingroup$
An interesting inequality, but even without the numerical evidence for the bound being around $3.16$ there doesn't seem to be any reason for $pi$ to feature here. Seems arbitrary
$endgroup$
– Yuriy S
Apr 25 '18 at 10:04
|
show 9 more comments
17
17
17
14
$begingroup$
If $a,b,c,d >0$, and $a+b+c+d=4$, prove that
$$a^{ab}+b^{bc}+c^{cd}+d^{da} geq pi.$$
I don't think Jensen's inequality will help here, but I think first determining where equality holds will be useful. Maybe taking the logarithm or exponential of both sides will also be useful, but I want to in the end get rid of the plus signs in order to simplify it.
inequality
asked May 6 '16 at 20:49
Puzzled417Puzzled417
3,105527
$endgroup$
If $a,b,c,d >0$, and $a+b+c+d=4$, prove that
$$a^{ab}+b^{bc}+c^{cd}+d^{da} geq pi.$$
I don't think Jensen's inequality will help here, but I think first determining where equality holds will be useful. Maybe taking the logarithm or exponential of both sides will also be useful, but I want to in the end get rid of the plus signs in order to simplify it.
inequality
inequality
asked May 6 '16 at 20:49
Puzzled417Puzzled417
3,105527
asked May 6 '16 at 20:49
Puzzled417Puzzled417
3,105527
asked May 6 '16 at 20:49
Puzzled417Puzzled417
3,105527
asked May 6 '16 at 20:49
Puzzled417Puzzled417
3,105527
asked May 6 '16 at 20:49
Puzzled417Puzzled417
3,105527
3,105527
$begingroup$
This reminds me the inequality $a^a+b^b>a^b+b^a$ from Wikipédia. Not sure if it help though...
$endgroup$
– Surb
May 6 '16 at 20:54
$begingroup$
May I ask how you came across this problem? Also, do you know whether the constant $pi$ is optimal?
$endgroup$
– Wojowu
May 6 '16 at 20:55
$begingroup$
@Wojowu It is from a problem solving website. Yes, I believe equality is achieved.
$endgroup$
– Puzzled417
May 6 '16 at 20:59
2
$begingroup$
The best I've got out of Mathematica is3.1605859174508652189…, usingNMinimize[{(a^2)^(a^2 b^2) + (b^2)^(b^2 c^2) + (c^2)^(c^2 d^2) + (d^2)^(d^2 a^2), a != 0 && b != 0 && c != 0 && d != 0 && a^2 + b^2 + c^2 + d^2 == 4}, {a, b, c, d}, WorkingPrecision -> 100]
$endgroup$
– Patrick Stevens
May 6 '16 at 21:26
2
$begingroup$
An interesting inequality, but even without the numerical evidence for the bound being around $3.16$ there doesn't seem to be any reason for $pi$ to feature here. Seems arbitrary
$endgroup$
– Yuriy S
Apr 25 '18 at 10:04
|
show 9 more comments
$begingroup$
This reminds me the inequality $a^a+b^b>a^b+b^a$ from Wikipédia. Not sure if it help though...
$endgroup$
– Surb
May 6 '16 at 20:54
$begingroup$
May I ask how you came across this problem? Also, do you know whether the constant $pi$ is optimal?
$endgroup$
– Wojowu
May 6 '16 at 20:55
$begingroup$
@Wojowu It is from a problem solving website. Yes, I believe equality is achieved.
$endgroup$
– Puzzled417
May 6 '16 at 20:59
2
$begingroup$
The best I've got out of Mathematica is3.1605859174508652189…, usingNMinimize[{(a^2)^(a^2 b^2) + (b^2)^(b^2 c^2) + (c^2)^(c^2 d^2) + (d^2)^(d^2 a^2), a != 0 && b != 0 && c != 0 && d != 0 && a^2 + b^2 + c^2 + d^2 == 4}, {a, b, c, d}, WorkingPrecision -> 100]
$endgroup$
– Patrick Stevens
May 6 '16 at 21:26
2
$begingroup$
An interesting inequality, but even without the numerical evidence for the bound being around $3.16$ there doesn't seem to be any reason for $pi$ to feature here. Seems arbitrary
$endgroup$
– Yuriy S
Apr 25 '18 at 10:04
$begingroup$
This reminds me the inequality $a^a+b^b>a^b+b^a$ from Wikipédia. Not sure if it help though...
$endgroup$
– Surb
May 6 '16 at 20:54
$begingroup$
This reminds me the inequality $a^a+b^b>a^b+b^a$ from Wikipédia. Not sure if it help though...
$endgroup$
– Surb
May 6 '16 at 20:54
$begingroup$
May I ask how you came across this problem? Also, do you know whether the constant $pi$ is optimal?
$endgroup$
– Wojowu
May 6 '16 at 20:55
$begingroup$
May I ask how you came across this problem? Also, do you know whether the constant $pi$ is optimal?
$endgroup$
– Wojowu
May 6 '16 at 20:55
$begingroup$
@Wojowu It is from a problem solving website. Yes, I believe equality is achieved.
$endgroup$
– Puzzled417
May 6 '16 at 20:59
$begingroup$
@Wojowu It is from a problem solving website. Yes, I believe equality is achieved.
$endgroup$
– Puzzled417
May 6 '16 at 20:59
2
2
$begingroup$
The best I've got out of Mathematica is
3.1605859174508652189…, using NMinimize[{(a^2)^(a^2 b^2) + (b^2)^(b^2 c^2) + (c^2)^(c^2 d^2) + (d^2)^(d^2 a^2), a != 0 && b != 0 && c != 0 && d != 0 && a^2 + b^2 + c^2 + d^2 == 4}, {a, b, c, d}, WorkingPrecision -> 100]$endgroup$
– Patrick Stevens
May 6 '16 at 21:26
$begingroup$
The best I've got out of Mathematica is
3.1605859174508652189…, using NMinimize[{(a^2)^(a^2 b^2) + (b^2)^(b^2 c^2) + (c^2)^(c^2 d^2) + (d^2)^(d^2 a^2), a != 0 && b != 0 && c != 0 && d != 0 && a^2 + b^2 + c^2 + d^2 == 4}, {a, b, c, d}, WorkingPrecision -> 100]$endgroup$
– Patrick Stevens
May 6 '16 at 21:26
2
2
$begingroup$
An interesting inequality, but even without the numerical evidence for the bound being around $3.16$ there doesn't seem to be any reason for $pi$ to feature here. Seems arbitrary
$endgroup$
– Yuriy S
Apr 25 '18 at 10:04
$begingroup$
An interesting inequality, but even without the numerical evidence for the bound being around $3.16$ there doesn't seem to be any reason for $pi$ to feature here. Seems arbitrary
$endgroup$
– Yuriy S
Apr 25 '18 at 10:04
|
show 9 more comments
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$begingroup$
This reminds me the inequality $a^a+b^b>a^b+b^a$ from Wikipédia. Not sure if it help though...
$endgroup$
– Surb
May 6 '16 at 20:54
$begingroup$
May I ask how you came across this problem? Also, do you know whether the constant $pi$ is optimal?
$endgroup$
– Wojowu
May 6 '16 at 20:55
$begingroup$
@Wojowu It is from a problem solving website. Yes, I believe equality is achieved.
$endgroup$
– Puzzled417
May 6 '16 at 20:59
2
$begingroup$
The best I've got out of Mathematica is
3.1605859174508652189…, usingNMinimize[{(a^2)^(a^2 b^2) + (b^2)^(b^2 c^2) + (c^2)^(c^2 d^2) + (d^2)^(d^2 a^2), a != 0 && b != 0 && c != 0 && d != 0 && a^2 + b^2 + c^2 + d^2 == 4}, {a, b, c, d}, WorkingPrecision -> 100]$endgroup$
– Patrick Stevens
May 6 '16 at 21:26
2
$begingroup$
An interesting inequality, but even without the numerical evidence for the bound being around $3.16$ there doesn't seem to be any reason for $pi$ to feature here. Seems arbitrary
$endgroup$
– Yuriy S
Apr 25 '18 at 10:04