Prove $mcos^2{theta} + nsin^2{theta} < l implies sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta} < sqrt{l} $...
$begingroup$
Prove that $mcos^2{theta} + nsin^2{theta} < l implies sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta} < sqrt{l} $ for every $m, n, l >0$.
calculus algebra-precalculus trigonometry inequality contest-math
edited Jan 19 at 18:47
Martin Sleziak
45.1k10123277
asked Sep 19 '14 at 15:30
user62029user62029
570523
$endgroup$
closed as off-topic by Did, metamorphy, Adrian Keister, clathratus, mrtaurho Jan 19 at 21:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, metamorphy, Adrian Keister, clathratus, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Prove that $mcos^2{theta} + nsin^2{theta} < l implies sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta} < sqrt{l} $ for every $m, n, l >0$.
calculus algebra-precalculus trigonometry inequality contest-math
edited Jan 19 at 18:47
Martin Sleziak
45.1k10123277
asked Sep 19 '14 at 15:30
user62029user62029
570523
$endgroup$
closed as off-topic by Did, metamorphy, Adrian Keister, clathratus, mrtaurho Jan 19 at 21:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, metamorphy, Adrian Keister, clathratus, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
is the hypothesis true for every $theta$?
$endgroup$
– mookid
Sep 19 '14 at 15:33
add a comment |
$begingroup$
Prove that $mcos^2{theta} + nsin^2{theta} < l implies sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta} < sqrt{l} $ for every $m, n, l >0$.
calculus algebra-precalculus trigonometry inequality contest-math
edited Jan 19 at 18:47
Martin Sleziak
45.1k10123277
asked Sep 19 '14 at 15:30
user62029user62029
570523
$endgroup$
Prove that $mcos^2{theta} + nsin^2{theta} < l implies sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta} < sqrt{l} $ for every $m, n, l >0$.
calculus algebra-precalculus trigonometry inequality contest-math
calculus algebra-precalculus trigonometry inequality contest-math
edited Jan 19 at 18:47
Martin Sleziak
45.1k10123277
asked Sep 19 '14 at 15:30
user62029user62029
570523
edited Jan 19 at 18:47
Martin Sleziak
45.1k10123277
asked Sep 19 '14 at 15:30
user62029user62029
570523
edited Jan 19 at 18:47
Martin Sleziak
45.1k10123277
edited Jan 19 at 18:47
Martin Sleziak
45.1k10123277
edited Jan 19 at 18:47
Martin Sleziak
45.1k10123277
45.1k10123277
asked Sep 19 '14 at 15:30
user62029user62029
570523
asked Sep 19 '14 at 15:30
user62029user62029
570523
asked Sep 19 '14 at 15:30
user62029user62029
570523
570523
closed as off-topic by Did, metamorphy, Adrian Keister, clathratus, mrtaurho Jan 19 at 21:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, metamorphy, Adrian Keister, clathratus, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, metamorphy, Adrian Keister, clathratus, mrtaurho Jan 19 at 21:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, metamorphy, Adrian Keister, clathratus, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
is the hypothesis true for every $theta$?
$endgroup$
– mookid
Sep 19 '14 at 15:33
add a comment |
1
$begingroup$
is the hypothesis true for every $theta$?
$endgroup$
– mookid
Sep 19 '14 at 15:33
1
1
$begingroup$
is the hypothesis true for every $theta$?
$endgroup$
– mookid
Sep 19 '14 at 15:33
$begingroup$
is the hypothesis true for every $theta$?
$endgroup$
– mookid
Sep 19 '14 at 15:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This problem can be found in
LARSON, LOREN C., 1983: Problem-Solving Through Problems.
Springer-Verlag, p. 255, 267.
The author proposes two solutions: the first one is based on the Cauchy-Schwarz inequality; the second one on the arithmetic-mean−geometric-mean inequality. A third solution, which uses the concavity of $ f(x) = sqrt{x} $, is left as an exercise.
I will write here the solution based on the second method:
begin{align*}
(sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta})^2
&= m cos^4{theta} + 2 sqrt{mn}cos^2{theta}sin^2{theta}+ nsin^4{theta} \
& leq m cos^4{theta} + (m + n)cos^2{theta}sin^2{theta}+ nsin^4{theta} \
&= (mcos^2{theta} + nsin^2{theta})(cos^2{theta} + sin^2{theta}) \
&= (mcos^2{theta} + nsin^2{theta})\
&<l.
end{align*}
$endgroup$
$begingroup$
Thank you for the reference and the hints. I'll try to solve the problem using all the methods.
$endgroup$
– user62029
Sep 19 '14 at 16:16
$begingroup$
You are welcome.
$endgroup$
– user161303
Sep 19 '14 at 16:22
add a comment |
$begingroup$
Hint: write
$$
sqrt{m}cos^2theta+sqrt{n}sin^2theta=(sqrt{m}costheta)cdotcostheta+(sqrt{n}sintheta)cdotsintheta
$$
and look into the Cauchy-Schwarz inequality.
answered Sep 19 '14 at 15:38
Kim Jong UnKim Jong Un
13.2k11738
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This problem can be found in
LARSON, LOREN C., 1983: Problem-Solving Through Problems.
Springer-Verlag, p. 255, 267.
The author proposes two solutions: the first one is based on the Cauchy-Schwarz inequality; the second one on the arithmetic-mean−geometric-mean inequality. A third solution, which uses the concavity of $ f(x) = sqrt{x} $, is left as an exercise.
I will write here the solution based on the second method:
begin{align*}
(sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta})^2
&= m cos^4{theta} + 2 sqrt{mn}cos^2{theta}sin^2{theta}+ nsin^4{theta} \
& leq m cos^4{theta} + (m + n)cos^2{theta}sin^2{theta}+ nsin^4{theta} \
&= (mcos^2{theta} + nsin^2{theta})(cos^2{theta} + sin^2{theta}) \
&= (mcos^2{theta} + nsin^2{theta})\
&<l.
end{align*}
$endgroup$
$begingroup$
Thank you for the reference and the hints. I'll try to solve the problem using all the methods.
$endgroup$
– user62029
Sep 19 '14 at 16:16
$begingroup$
You are welcome.
$endgroup$
– user161303
Sep 19 '14 at 16:22
add a comment |
$begingroup$
This problem can be found in
LARSON, LOREN C., 1983: Problem-Solving Through Problems.
Springer-Verlag, p. 255, 267.
The author proposes two solutions: the first one is based on the Cauchy-Schwarz inequality; the second one on the arithmetic-mean−geometric-mean inequality. A third solution, which uses the concavity of $ f(x) = sqrt{x} $, is left as an exercise.
I will write here the solution based on the second method:
begin{align*}
(sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta})^2
&= m cos^4{theta} + 2 sqrt{mn}cos^2{theta}sin^2{theta}+ nsin^4{theta} \
& leq m cos^4{theta} + (m + n)cos^2{theta}sin^2{theta}+ nsin^4{theta} \
&= (mcos^2{theta} + nsin^2{theta})(cos^2{theta} + sin^2{theta}) \
&= (mcos^2{theta} + nsin^2{theta})\
&<l.
end{align*}
$endgroup$
$begingroup$
Thank you for the reference and the hints. I'll try to solve the problem using all the methods.
$endgroup$
– user62029
Sep 19 '14 at 16:16
$begingroup$
You are welcome.
$endgroup$
– user161303
Sep 19 '14 at 16:22
add a comment |
$begingroup$
This problem can be found in
LARSON, LOREN C., 1983: Problem-Solving Through Problems.
Springer-Verlag, p. 255, 267.
The author proposes two solutions: the first one is based on the Cauchy-Schwarz inequality; the second one on the arithmetic-mean−geometric-mean inequality. A third solution, which uses the concavity of $ f(x) = sqrt{x} $, is left as an exercise.
I will write here the solution based on the second method:
begin{align*}
(sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta})^2
&= m cos^4{theta} + 2 sqrt{mn}cos^2{theta}sin^2{theta}+ nsin^4{theta} \
& leq m cos^4{theta} + (m + n)cos^2{theta}sin^2{theta}+ nsin^4{theta} \
&= (mcos^2{theta} + nsin^2{theta})(cos^2{theta} + sin^2{theta}) \
&= (mcos^2{theta} + nsin^2{theta})\
&<l.
end{align*}
$endgroup$
This problem can be found in
LARSON, LOREN C., 1983: Problem-Solving Through Problems.
Springer-Verlag, p. 255, 267.
The author proposes two solutions: the first one is based on the Cauchy-Schwarz inequality; the second one on the arithmetic-mean−geometric-mean inequality. A third solution, which uses the concavity of $ f(x) = sqrt{x} $, is left as an exercise.
I will write here the solution based on the second method:
begin{align*}
(sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta})^2
&= m cos^4{theta} + 2 sqrt{mn}cos^2{theta}sin^2{theta}+ nsin^4{theta} \
& leq m cos^4{theta} + (m + n)cos^2{theta}sin^2{theta}+ nsin^4{theta} \
&= (mcos^2{theta} + nsin^2{theta})(cos^2{theta} + sin^2{theta}) \
&= (mcos^2{theta} + nsin^2{theta})\
&<l.
end{align*}
answered Sep 19 '14 at 15:52
user161303
$begingroup$
Thank you for the reference and the hints. I'll try to solve the problem using all the methods.
$endgroup$
– user62029
Sep 19 '14 at 16:16
$begingroup$
You are welcome.
$endgroup$
– user161303
Sep 19 '14 at 16:22
add a comment |
$begingroup$
Thank you for the reference and the hints. I'll try to solve the problem using all the methods.
$endgroup$
– user62029
Sep 19 '14 at 16:16
$begingroup$
You are welcome.
$endgroup$
– user161303
Sep 19 '14 at 16:22
$begingroup$
Thank you for the reference and the hints. I'll try to solve the problem using all the methods.
$endgroup$
– user62029
Sep 19 '14 at 16:16
$begingroup$
Thank you for the reference and the hints. I'll try to solve the problem using all the methods.
$endgroup$
– user62029
Sep 19 '14 at 16:16
$begingroup$
You are welcome.
$endgroup$
– user161303
Sep 19 '14 at 16:22
$begingroup$
You are welcome.
$endgroup$
– user161303
Sep 19 '14 at 16:22
add a comment |
$begingroup$
Hint: write
$$
sqrt{m}cos^2theta+sqrt{n}sin^2theta=(sqrt{m}costheta)cdotcostheta+(sqrt{n}sintheta)cdotsintheta
$$
and look into the Cauchy-Schwarz inequality.
answered Sep 19 '14 at 15:38
Kim Jong UnKim Jong Un
13.2k11738
$endgroup$
add a comment |
$begingroup$
Hint: write
$$
sqrt{m}cos^2theta+sqrt{n}sin^2theta=(sqrt{m}costheta)cdotcostheta+(sqrt{n}sintheta)cdotsintheta
$$
and look into the Cauchy-Schwarz inequality.
answered Sep 19 '14 at 15:38
Kim Jong UnKim Jong Un
13.2k11738
$endgroup$
add a comment |
$begingroup$
Hint: write
$$
sqrt{m}cos^2theta+sqrt{n}sin^2theta=(sqrt{m}costheta)cdotcostheta+(sqrt{n}sintheta)cdotsintheta
$$
and look into the Cauchy-Schwarz inequality.
answered Sep 19 '14 at 15:38
Kim Jong UnKim Jong Un
13.2k11738
$endgroup$
Hint: write
$$
sqrt{m}cos^2theta+sqrt{n}sin^2theta=(sqrt{m}costheta)cdotcostheta+(sqrt{n}sintheta)cdotsintheta
$$
and look into the Cauchy-Schwarz inequality.
answered Sep 19 '14 at 15:38
Kim Jong UnKim Jong Un
13.2k11738
answered Sep 19 '14 at 15:38
Kim Jong UnKim Jong Un
13.2k11738
answered Sep 19 '14 at 15:38
Kim Jong UnKim Jong Un
13.2k11738
answered Sep 19 '14 at 15:38
Kim Jong UnKim Jong Un
13.2k11738
13.2k11738
add a comment |
add a comment |
1
$begingroup$
is the hypothesis true for every $theta$?
$endgroup$
– mookid
Sep 19 '14 at 15:33