If $frac{a}{b}<frac{a'}{b'}<frac{a''}{b''}$, with all values positive,...
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This question already has an answer here:
Inequality from Challenge and Thrill of Pre-college Mathematics [closed]
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Bonjour. I need help for this.
Let $a$, $b$, $a^prime$, $b^prime$, $a^{primeprime}$, and $b^{primeprime}$ be six non-zero positive numbers such that
$$frac{a}{b}<frac{a^prime}{b^prime}<frac{a^{primeprime}}{b^{primeprime}}$$
We want to prove that this stuff holds:
$$frac{a}{b}<frac{ab+a^prime b^prime+a^{primeprime}b^{primeprime}}{b^2+{b^{prime}}^2+{b^{primeprime}}^2}<frac{a^{primeprime}}{b^{primeprime}}$$
Thanks.
inequality
edited Jan 13 at 14:24
Blue
49.1k870156
asked Jan 13 at 14:11
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,13929
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marked as duplicate by RRL, Martin R, Misha Lavrov, max_zorn, Cesareo Jan 27 at 1:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Inequality from Challenge and Thrill of Pre-college Mathematics [closed]
1 answer
Bonjour. I need help for this.
Let $a$, $b$, $a^prime$, $b^prime$, $a^{primeprime}$, and $b^{primeprime}$ be six non-zero positive numbers such that
$$frac{a}{b}<frac{a^prime}{b^prime}<frac{a^{primeprime}}{b^{primeprime}}$$
We want to prove that this stuff holds:
$$frac{a}{b}<frac{ab+a^prime b^prime+a^{primeprime}b^{primeprime}}{b^2+{b^{prime}}^2+{b^{primeprime}}^2}<frac{a^{primeprime}}{b^{primeprime}}$$
Thanks.
inequality
edited Jan 13 at 14:24
Blue
49.1k870156
asked Jan 13 at 14:11
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,13929
$endgroup$
marked as duplicate by RRL, Martin R, Misha Lavrov, max_zorn, Cesareo Jan 27 at 1:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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what have you tried so far?
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– Siong Thye Goh
Jan 13 at 14:18
add a comment |
$begingroup$
This question already has an answer here:
Inequality from Challenge and Thrill of Pre-college Mathematics [closed]
1 answer
Bonjour. I need help for this.
Let $a$, $b$, $a^prime$, $b^prime$, $a^{primeprime}$, and $b^{primeprime}$ be six non-zero positive numbers such that
$$frac{a}{b}<frac{a^prime}{b^prime}<frac{a^{primeprime}}{b^{primeprime}}$$
We want to prove that this stuff holds:
$$frac{a}{b}<frac{ab+a^prime b^prime+a^{primeprime}b^{primeprime}}{b^2+{b^{prime}}^2+{b^{primeprime}}^2}<frac{a^{primeprime}}{b^{primeprime}}$$
Thanks.
inequality
edited Jan 13 at 14:24
Blue
49.1k870156
asked Jan 13 at 14:11
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,13929
$endgroup$
This question already has an answer here:
Inequality from Challenge and Thrill of Pre-college Mathematics [closed]
1 answer
Bonjour. I need help for this.
Let $a$, $b$, $a^prime$, $b^prime$, $a^{primeprime}$, and $b^{primeprime}$ be six non-zero positive numbers such that
$$frac{a}{b}<frac{a^prime}{b^prime}<frac{a^{primeprime}}{b^{primeprime}}$$
We want to prove that this stuff holds:
$$frac{a}{b}<frac{ab+a^prime b^prime+a^{primeprime}b^{primeprime}}{b^2+{b^{prime}}^2+{b^{primeprime}}^2}<frac{a^{primeprime}}{b^{primeprime}}$$
Thanks.
This question already has an answer here:
Inequality from Challenge and Thrill of Pre-college Mathematics [closed]
1 answer
inequality
inequality
edited Jan 13 at 14:24
Blue
49.1k870156
asked Jan 13 at 14:11
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,13929
edited Jan 13 at 14:24
Blue
49.1k870156
asked Jan 13 at 14:11
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,13929
edited Jan 13 at 14:24
Blue
49.1k870156
edited Jan 13 at 14:24
Blue
49.1k870156
edited Jan 13 at 14:24
Blue
49.1k870156
49.1k870156
asked Jan 13 at 14:11
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,13929
asked Jan 13 at 14:11
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,13929
asked Jan 13 at 14:11
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,13929
1,13929
marked as duplicate by RRL, Martin R, Misha Lavrov, max_zorn, Cesareo Jan 27 at 1:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by RRL, Martin R, Misha Lavrov, max_zorn, Cesareo Jan 27 at 1:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
what have you tried so far?
$endgroup$
– Siong Thye Goh
Jan 13 at 14:18
add a comment |
$begingroup$
what have you tried so far?
$endgroup$
– Siong Thye Goh
Jan 13 at 14:18
$begingroup$
what have you tried so far?
$endgroup$
– Siong Thye Goh
Jan 13 at 14:18
$begingroup$
what have you tried so far?
$endgroup$
– Siong Thye Goh
Jan 13 at 14:18
add a comment |
2 Answers
2
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Let $frac{a}{b}=k,$ $frac{a'}{b'}=m$ and $frac{a''}{b''}=n$,$b=x$, $b'=y$ and $b''=z$.
Thus, $k<m<n$ and we need to prove that
$$k<frac{kx^2+my^2+nz^2}{x^2+y^2+z^2}<n,$$ which is true because the left inequality it's
$$(m-k)y^2+(n-k)z^2>0$$ and the right inequality it's
$$(n-m)y^2+(n-k)x^2>0.$$
answered Jan 13 at 14:25
Michael RozenbergMichael Rozenberg
108k1895200
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We multiply the left inequality by $$b^2+b_1^2+b_2^2>0$$ then we get
$$a_1b_1b+a_2b_2b>ab_1^2+ab_2^2$$ and this is
$$b_1(a_1b-ab_1)+b_2(a_2b-ab_2)>0$$ this is true since we have
$$frac{a_1}{b_1}>frac{a}{b}$$ and $$frac{a_2}{b_2}>frac{a}{b}$$
answered Jan 13 at 14:27
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $frac{a}{b}=k,$ $frac{a'}{b'}=m$ and $frac{a''}{b''}=n$,$b=x$, $b'=y$ and $b''=z$.
Thus, $k<m<n$ and we need to prove that
$$k<frac{kx^2+my^2+nz^2}{x^2+y^2+z^2}<n,$$ which is true because the left inequality it's
$$(m-k)y^2+(n-k)z^2>0$$ and the right inequality it's
$$(n-m)y^2+(n-k)x^2>0.$$
answered Jan 13 at 14:25
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
Let $frac{a}{b}=k,$ $frac{a'}{b'}=m$ and $frac{a''}{b''}=n$,$b=x$, $b'=y$ and $b''=z$.
Thus, $k<m<n$ and we need to prove that
$$k<frac{kx^2+my^2+nz^2}{x^2+y^2+z^2}<n,$$ which is true because the left inequality it's
$$(m-k)y^2+(n-k)z^2>0$$ and the right inequality it's
$$(n-m)y^2+(n-k)x^2>0.$$
answered Jan 13 at 14:25
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
Let $frac{a}{b}=k,$ $frac{a'}{b'}=m$ and $frac{a''}{b''}=n$,$b=x$, $b'=y$ and $b''=z$.
Thus, $k<m<n$ and we need to prove that
$$k<frac{kx^2+my^2+nz^2}{x^2+y^2+z^2}<n,$$ which is true because the left inequality it's
$$(m-k)y^2+(n-k)z^2>0$$ and the right inequality it's
$$(n-m)y^2+(n-k)x^2>0.$$
answered Jan 13 at 14:25
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
Let $frac{a}{b}=k,$ $frac{a'}{b'}=m$ and $frac{a''}{b''}=n$,$b=x$, $b'=y$ and $b''=z$.
Thus, $k<m<n$ and we need to prove that
$$k<frac{kx^2+my^2+nz^2}{x^2+y^2+z^2}<n,$$ which is true because the left inequality it's
$$(m-k)y^2+(n-k)z^2>0$$ and the right inequality it's
$$(n-m)y^2+(n-k)x^2>0.$$
answered Jan 13 at 14:25
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 13 at 14:25
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 13 at 14:25
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 13 at 14:25
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
add a comment |
add a comment |
$begingroup$
We multiply the left inequality by $$b^2+b_1^2+b_2^2>0$$ then we get
$$a_1b_1b+a_2b_2b>ab_1^2+ab_2^2$$ and this is
$$b_1(a_1b-ab_1)+b_2(a_2b-ab_2)>0$$ this is true since we have
$$frac{a_1}{b_1}>frac{a}{b}$$ and $$frac{a_2}{b_2}>frac{a}{b}$$
answered Jan 13 at 14:27
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
add a comment |
$begingroup$
We multiply the left inequality by $$b^2+b_1^2+b_2^2>0$$ then we get
$$a_1b_1b+a_2b_2b>ab_1^2+ab_2^2$$ and this is
$$b_1(a_1b-ab_1)+b_2(a_2b-ab_2)>0$$ this is true since we have
$$frac{a_1}{b_1}>frac{a}{b}$$ and $$frac{a_2}{b_2}>frac{a}{b}$$
answered Jan 13 at 14:27
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
add a comment |
$begingroup$
We multiply the left inequality by $$b^2+b_1^2+b_2^2>0$$ then we get
$$a_1b_1b+a_2b_2b>ab_1^2+ab_2^2$$ and this is
$$b_1(a_1b-ab_1)+b_2(a_2b-ab_2)>0$$ this is true since we have
$$frac{a_1}{b_1}>frac{a}{b}$$ and $$frac{a_2}{b_2}>frac{a}{b}$$
answered Jan 13 at 14:27
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
We multiply the left inequality by $$b^2+b_1^2+b_2^2>0$$ then we get
$$a_1b_1b+a_2b_2b>ab_1^2+ab_2^2$$ and this is
$$b_1(a_1b-ab_1)+b_2(a_2b-ab_2)>0$$ this is true since we have
$$frac{a_1}{b_1}>frac{a}{b}$$ and $$frac{a_2}{b_2}>frac{a}{b}$$
answered Jan 13 at 14:27
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
answered Jan 13 at 14:27
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
answered Jan 13 at 14:27
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
answered Jan 13 at 14:27
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
77.9k42866
add a comment |
add a comment |
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$begingroup$
what have you tried so far?
$endgroup$
– Siong Thye Goh
Jan 13 at 14:18