Find the least positive real number $k$ such that $7sqrt{a} + 17sqrt{b} + ksqrt{c} ge sqrt{2019}$ over all...
$begingroup$
Working on a problem...
Find the least positive real number $k$ such that $7sqrt{a} + 17sqrt{b} + ksqrt{c} ge sqrt{2019}$ over all positive real numbers $a,b,c$ with $a+b+c=1$.
Maximizing the "$a$" term doesn't seem to work, and expansion through moving a radical to the right side of the equation simply leads to more radicals.
Any help?
contest-math
edited Jan 10 at 6:49
Thomas Shelby
2,408221
asked Jan 10 at 5:55
Hubert LioniHubert Lioni
133
$endgroup$
add a comment |
$begingroup$
Working on a problem...
Find the least positive real number $k$ such that $7sqrt{a} + 17sqrt{b} + ksqrt{c} ge sqrt{2019}$ over all positive real numbers $a,b,c$ with $a+b+c=1$.
Maximizing the "$a$" term doesn't seem to work, and expansion through moving a radical to the right side of the equation simply leads to more radicals.
Any help?
contest-math
edited Jan 10 at 6:49
Thomas Shelby
2,408221
asked Jan 10 at 5:55
Hubert LioniHubert Lioni
133
$endgroup$
$begingroup$
I would just write $k=frac {sqrt{2019}-7sqrt a-17sqrt b}{sqrt {1-a-b}}$ and take derivatives with respect to $a$ and $b$, set to zero, and try to solve the simultaneous equations.
$endgroup$
– Ross Millikan
Jan 10 at 6:11
add a comment |
$begingroup$
Working on a problem...
Find the least positive real number $k$ such that $7sqrt{a} + 17sqrt{b} + ksqrt{c} ge sqrt{2019}$ over all positive real numbers $a,b,c$ with $a+b+c=1$.
Maximizing the "$a$" term doesn't seem to work, and expansion through moving a radical to the right side of the equation simply leads to more radicals.
Any help?
contest-math
edited Jan 10 at 6:49
Thomas Shelby
2,408221
asked Jan 10 at 5:55
Hubert LioniHubert Lioni
133
$endgroup$
Working on a problem...
Find the least positive real number $k$ such that $7sqrt{a} + 17sqrt{b} + ksqrt{c} ge sqrt{2019}$ over all positive real numbers $a,b,c$ with $a+b+c=1$.
Maximizing the "$a$" term doesn't seem to work, and expansion through moving a radical to the right side of the equation simply leads to more radicals.
Any help?
contest-math
contest-math
edited Jan 10 at 6:49
Thomas Shelby
2,408221
asked Jan 10 at 5:55
Hubert LioniHubert Lioni
133
edited Jan 10 at 6:49
Thomas Shelby
2,408221
asked Jan 10 at 5:55
Hubert LioniHubert Lioni
133
edited Jan 10 at 6:49
Thomas Shelby
2,408221
edited Jan 10 at 6:49
Thomas Shelby
2,408221
edited Jan 10 at 6:49
Thomas Shelby
2,408221
2,408221
asked Jan 10 at 5:55
Hubert LioniHubert Lioni
133
asked Jan 10 at 5:55
Hubert LioniHubert Lioni
133
asked Jan 10 at 5:55
Hubert LioniHubert Lioni
133
133
$begingroup$
I would just write $k=frac {sqrt{2019}-7sqrt a-17sqrt b}{sqrt {1-a-b}}$ and take derivatives with respect to $a$ and $b$, set to zero, and try to solve the simultaneous equations.
$endgroup$
– Ross Millikan
Jan 10 at 6:11
add a comment |
$begingroup$
I would just write $k=frac {sqrt{2019}-7sqrt a-17sqrt b}{sqrt {1-a-b}}$ and take derivatives with respect to $a$ and $b$, set to zero, and try to solve the simultaneous equations.
$endgroup$
– Ross Millikan
Jan 10 at 6:11
$begingroup$
I would just write $k=frac {sqrt{2019}-7sqrt a-17sqrt b}{sqrt {1-a-b}}$ and take derivatives with respect to $a$ and $b$, set to zero, and try to solve the simultaneous equations.
$endgroup$
– Ross Millikan
Jan 10 at 6:11
$begingroup$
I would just write $k=frac {sqrt{2019}-7sqrt a-17sqrt b}{sqrt {1-a-b}}$ and take derivatives with respect to $a$ and $b$, set to zero, and try to solve the simultaneous equations.
$endgroup$
– Ross Millikan
Jan 10 at 6:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This response is informal handwaving, which I suspect leads to the accurate answer. Any value apportioned to a, or b, is deducted from c.
Since $sqrt{2019} > 17, k$ must be greater than 17. Therefore the solution lies in maximizing c and setting k correspondingly. With a=0=b, and c=1, then $k=sqrt{2019}.$
I will be very surprised if this informal approach is wrong.
Addendum
It occurs to me that I may have misinterpreted the problem. If it is desired to identify K so that the inequality holds regardless of how a,b,c are apportioned (as long as a,b,c positive, a+b+c=1) then there is no answer. Regardless of the fixed value of k, as c approaches 0,
$ksqrt{c}$ will approach 0, and the left hand side will then be no greater than 17.
Addendum 2
Are you sure that you have the constraints right?
answered Jan 10 at 6:59
user2661923user2661923
510112
$endgroup$
add a comment |
$begingroup$
It cannot be done. Put
$$a=1-2epsilon^2,quad b=c=epsilon^2$$
with $0<epsilonll1$. Then we want
$$7sqrt{1-2epsilon^2}+(k+17)epsilongeqsqrt{2019} ,$$
or
$$(k+17)epsilongeq44.933-7sqrt{1-2epsilon^2}>37 .$$
There is no $k$ that can do this for all $epsilon>0$.
answered Jan 10 at 14:16
Christian BlatterChristian Blatter
172k7113326
$endgroup$
add a comment |
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2 Answers
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2 Answers
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votes
$begingroup$
This response is informal handwaving, which I suspect leads to the accurate answer. Any value apportioned to a, or b, is deducted from c.
Since $sqrt{2019} > 17, k$ must be greater than 17. Therefore the solution lies in maximizing c and setting k correspondingly. With a=0=b, and c=1, then $k=sqrt{2019}.$
I will be very surprised if this informal approach is wrong.
Addendum
It occurs to me that I may have misinterpreted the problem. If it is desired to identify K so that the inequality holds regardless of how a,b,c are apportioned (as long as a,b,c positive, a+b+c=1) then there is no answer. Regardless of the fixed value of k, as c approaches 0,
$ksqrt{c}$ will approach 0, and the left hand side will then be no greater than 17.
Addendum 2
Are you sure that you have the constraints right?
answered Jan 10 at 6:59
user2661923user2661923
510112
$endgroup$
add a comment |
$begingroup$
This response is informal handwaving, which I suspect leads to the accurate answer. Any value apportioned to a, or b, is deducted from c.
Since $sqrt{2019} > 17, k$ must be greater than 17. Therefore the solution lies in maximizing c and setting k correspondingly. With a=0=b, and c=1, then $k=sqrt{2019}.$
I will be very surprised if this informal approach is wrong.
Addendum
It occurs to me that I may have misinterpreted the problem. If it is desired to identify K so that the inequality holds regardless of how a,b,c are apportioned (as long as a,b,c positive, a+b+c=1) then there is no answer. Regardless of the fixed value of k, as c approaches 0,
$ksqrt{c}$ will approach 0, and the left hand side will then be no greater than 17.
Addendum 2
Are you sure that you have the constraints right?
answered Jan 10 at 6:59
user2661923user2661923
510112
$endgroup$
add a comment |
$begingroup$
This response is informal handwaving, which I suspect leads to the accurate answer. Any value apportioned to a, or b, is deducted from c.
Since $sqrt{2019} > 17, k$ must be greater than 17. Therefore the solution lies in maximizing c and setting k correspondingly. With a=0=b, and c=1, then $k=sqrt{2019}.$
I will be very surprised if this informal approach is wrong.
Addendum
It occurs to me that I may have misinterpreted the problem. If it is desired to identify K so that the inequality holds regardless of how a,b,c are apportioned (as long as a,b,c positive, a+b+c=1) then there is no answer. Regardless of the fixed value of k, as c approaches 0,
$ksqrt{c}$ will approach 0, and the left hand side will then be no greater than 17.
Addendum 2
Are you sure that you have the constraints right?
answered Jan 10 at 6:59
user2661923user2661923
510112
$endgroup$
This response is informal handwaving, which I suspect leads to the accurate answer. Any value apportioned to a, or b, is deducted from c.
Since $sqrt{2019} > 17, k$ must be greater than 17. Therefore the solution lies in maximizing c and setting k correspondingly. With a=0=b, and c=1, then $k=sqrt{2019}.$
I will be very surprised if this informal approach is wrong.
Addendum
It occurs to me that I may have misinterpreted the problem. If it is desired to identify K so that the inequality holds regardless of how a,b,c are apportioned (as long as a,b,c positive, a+b+c=1) then there is no answer. Regardless of the fixed value of k, as c approaches 0,
$ksqrt{c}$ will approach 0, and the left hand side will then be no greater than 17.
Addendum 2
Are you sure that you have the constraints right?
answered Jan 10 at 6:59
user2661923user2661923
510112
edited Jan 10 at 7:04
answered Jan 10 at 6:59
user2661923user2661923
510112
answered Jan 10 at 6:59
user2661923user2661923
510112
answered Jan 10 at 6:59
user2661923user2661923
510112
510112
add a comment |
add a comment |
$begingroup$
It cannot be done. Put
$$a=1-2epsilon^2,quad b=c=epsilon^2$$
with $0<epsilonll1$. Then we want
$$7sqrt{1-2epsilon^2}+(k+17)epsilongeqsqrt{2019} ,$$
or
$$(k+17)epsilongeq44.933-7sqrt{1-2epsilon^2}>37 .$$
There is no $k$ that can do this for all $epsilon>0$.
answered Jan 10 at 14:16
Christian BlatterChristian Blatter
172k7113326
$endgroup$
add a comment |
$begingroup$
It cannot be done. Put
$$a=1-2epsilon^2,quad b=c=epsilon^2$$
with $0<epsilonll1$. Then we want
$$7sqrt{1-2epsilon^2}+(k+17)epsilongeqsqrt{2019} ,$$
or
$$(k+17)epsilongeq44.933-7sqrt{1-2epsilon^2}>37 .$$
There is no $k$ that can do this for all $epsilon>0$.
answered Jan 10 at 14:16
Christian BlatterChristian Blatter
172k7113326
$endgroup$
add a comment |
$begingroup$
It cannot be done. Put
$$a=1-2epsilon^2,quad b=c=epsilon^2$$
with $0<epsilonll1$. Then we want
$$7sqrt{1-2epsilon^2}+(k+17)epsilongeqsqrt{2019} ,$$
or
$$(k+17)epsilongeq44.933-7sqrt{1-2epsilon^2}>37 .$$
There is no $k$ that can do this for all $epsilon>0$.
answered Jan 10 at 14:16
Christian BlatterChristian Blatter
172k7113326
$endgroup$
It cannot be done. Put
$$a=1-2epsilon^2,quad b=c=epsilon^2$$
with $0<epsilonll1$. Then we want
$$7sqrt{1-2epsilon^2}+(k+17)epsilongeqsqrt{2019} ,$$
or
$$(k+17)epsilongeq44.933-7sqrt{1-2epsilon^2}>37 .$$
There is no $k$ that can do this for all $epsilon>0$.
answered Jan 10 at 14:16
Christian BlatterChristian Blatter
172k7113326
answered Jan 10 at 14:16
Christian BlatterChristian Blatter
172k7113326
answered Jan 10 at 14:16
Christian BlatterChristian Blatter
172k7113326
answered Jan 10 at 14:16
Christian BlatterChristian Blatter
172k7113326
172k7113326
add a comment |
add a comment |
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$begingroup$
I would just write $k=frac {sqrt{2019}-7sqrt a-17sqrt b}{sqrt {1-a-b}}$ and take derivatives with respect to $a$ and $b$, set to zero, and try to solve the simultaneous equations.
$endgroup$
– Ross Millikan
Jan 10 at 6:11