Mixing
How do I prove this relationship between positive terms of a G.P.?
$begingroup$
$a$, $b$, $c$, and $d$ are positive terms of a G.P. This is the relationship I'm trying to prove:
$$frac1{ab} + frac1{cd} > 2 left(frac1{bd} + frac1{ac} - frac1{ad}right)$$
This question was listed under the section on Arithmetic, Geometric, and Harmonic Means. So, I tried using those.
$$A = frac1{ab}$$
$$B = frac1{cd}$$
$$A.M. = frac12(A + B) = frac12left(frac1{ab}+frac1{cd}right)$$
This gives the correct term on the left side as well as the $2$ on the right side.
But I am completely blank from here. How do I proceed from here to prove this relationship using means?
sequences-and-series inequality arithmetic-progressions geometric-series geometric-progressions
edited Jan 29 at 14:47
Maria Mazur
49.3k1360123
asked Jan 29 at 13:41
user638500user638500
403
$endgroup$
add a comment |
$begingroup$
$a$, $b$, $c$, and $d$ are positive terms of a G.P. This is the relationship I'm trying to prove:
$$frac1{ab} + frac1{cd} > 2 left(frac1{bd} + frac1{ac} - frac1{ad}right)$$
This question was listed under the section on Arithmetic, Geometric, and Harmonic Means. So, I tried using those.
$$A = frac1{ab}$$
$$B = frac1{cd}$$
$$A.M. = frac12(A + B) = frac12left(frac1{ab}+frac1{cd}right)$$
This gives the correct term on the left side as well as the $2$ on the right side.
But I am completely blank from here. How do I proceed from here to prove this relationship using means?
sequences-and-series inequality arithmetic-progressions geometric-series geometric-progressions
edited Jan 29 at 14:47
Maria Mazur
49.3k1360123
asked Jan 29 at 13:41
user638500user638500
403
$endgroup$
add a comment |
$begingroup$
$a$, $b$, $c$, and $d$ are positive terms of a G.P. This is the relationship I'm trying to prove:
$$frac1{ab} + frac1{cd} > 2 left(frac1{bd} + frac1{ac} - frac1{ad}right)$$
This question was listed under the section on Arithmetic, Geometric, and Harmonic Means. So, I tried using those.
$$A = frac1{ab}$$
$$B = frac1{cd}$$
$$A.M. = frac12(A + B) = frac12left(frac1{ab}+frac1{cd}right)$$
This gives the correct term on the left side as well as the $2$ on the right side.
But I am completely blank from here. How do I proceed from here to prove this relationship using means?
sequences-and-series inequality arithmetic-progressions geometric-series geometric-progressions
edited Jan 29 at 14:47
Maria Mazur
49.3k1360123
asked Jan 29 at 13:41
user638500user638500
403
$endgroup$
$a$, $b$, $c$, and $d$ are positive terms of a G.P. This is the relationship I'm trying to prove:
$$frac1{ab} + frac1{cd} > 2 left(frac1{bd} + frac1{ac} - frac1{ad}right)$$
This question was listed under the section on Arithmetic, Geometric, and Harmonic Means. So, I tried using those.
$$A = frac1{ab}$$
$$B = frac1{cd}$$
$$A.M. = frac12(A + B) = frac12left(frac1{ab}+frac1{cd}right)$$
This gives the correct term on the left side as well as the $2$ on the right side.
But I am completely blank from here. How do I proceed from here to prove this relationship using means?
sequences-and-series inequality arithmetic-progressions geometric-series geometric-progressions
sequences-and-series inequality arithmetic-progressions geometric-series geometric-progressions
edited Jan 29 at 14:47
Maria Mazur
49.3k1360123
asked Jan 29 at 13:41
user638500user638500
403
edited Jan 29 at 14:47
Maria Mazur
49.3k1360123
asked Jan 29 at 13:41
user638500user638500
403
edited Jan 29 at 14:47
Maria Mazur
49.3k1360123
edited Jan 29 at 14:47
Maria Mazur
49.3k1360123
edited Jan 29 at 14:47
Maria Mazur
49.3k1360123
49.3k1360123
asked Jan 29 at 13:41
user638500user638500
403
asked Jan 29 at 13:41
user638500user638500
403
asked Jan 29 at 13:41
user638500user638500
403
403
add a comment |
add a comment |
1 Answer
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$begingroup$
So $b=ax, c=ax^2$, and $d=ax^3$, and you have to prove: $$frac1{a^2x} + frac1{a^2x^5} > 2 (frac1{a^2x^4} + frac1{a^2x^2} - frac1{a^2x^3})$$
or $$1 + frac1{x^4} > 2 (frac1{x^3} + frac1{x} - frac1{x^2})$$
or $$boxed{x^4+1>2(x+x^3-x^2)}$$
or
$$x^4-2x^3+2x^2-2x+1>0iff (x^2-x)^2+(x-1)^2>0,$$ which is obviously true.
You can try to prove the boxed inequality also like this, using AM-GM inequality: $$x^4+x^2geq 2sqrt{x^4cdot x^2} = 2x^3$$ and$$x^2+1geq 2sqrt{x^2cdot 1} = 2x$$
so $$x^4+2x^2+x = (x^4+x^2)+(x^2+1)> 2x^3+2x$$ and we are done again.
edited Jan 29 at 14:45
J. W. Tanner
4,1961320
answered Jan 29 at 13:52
Maria MazurMaria Mazur
49.3k1360123
$endgroup$
$begingroup$
Sorry, I am terrible at maths. Can you please hint on how to proceed further?
$endgroup$
– user638500
Jan 29 at 14:07
add a comment |
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1 Answer
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1 Answer
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$begingroup$
So $b=ax, c=ax^2$, and $d=ax^3$, and you have to prove: $$frac1{a^2x} + frac1{a^2x^5} > 2 (frac1{a^2x^4} + frac1{a^2x^2} - frac1{a^2x^3})$$
or $$1 + frac1{x^4} > 2 (frac1{x^3} + frac1{x} - frac1{x^2})$$
or $$boxed{x^4+1>2(x+x^3-x^2)}$$
or
$$x^4-2x^3+2x^2-2x+1>0iff (x^2-x)^2+(x-1)^2>0,$$ which is obviously true.
You can try to prove the boxed inequality also like this, using AM-GM inequality: $$x^4+x^2geq 2sqrt{x^4cdot x^2} = 2x^3$$ and$$x^2+1geq 2sqrt{x^2cdot 1} = 2x$$
so $$x^4+2x^2+x = (x^4+x^2)+(x^2+1)> 2x^3+2x$$ and we are done again.
edited Jan 29 at 14:45
J. W. Tanner
4,1961320
answered Jan 29 at 13:52
Maria MazurMaria Mazur
49.3k1360123
$endgroup$
$begingroup$
Sorry, I am terrible at maths. Can you please hint on how to proceed further?
$endgroup$
– user638500
Jan 29 at 14:07
add a comment |
$begingroup$
So $b=ax, c=ax^2$, and $d=ax^3$, and you have to prove: $$frac1{a^2x} + frac1{a^2x^5} > 2 (frac1{a^2x^4} + frac1{a^2x^2} - frac1{a^2x^3})$$
or $$1 + frac1{x^4} > 2 (frac1{x^3} + frac1{x} - frac1{x^2})$$
or $$boxed{x^4+1>2(x+x^3-x^2)}$$
or
$$x^4-2x^3+2x^2-2x+1>0iff (x^2-x)^2+(x-1)^2>0,$$ which is obviously true.
You can try to prove the boxed inequality also like this, using AM-GM inequality: $$x^4+x^2geq 2sqrt{x^4cdot x^2} = 2x^3$$ and$$x^2+1geq 2sqrt{x^2cdot 1} = 2x$$
so $$x^4+2x^2+x = (x^4+x^2)+(x^2+1)> 2x^3+2x$$ and we are done again.
edited Jan 29 at 14:45
J. W. Tanner
4,1961320
answered Jan 29 at 13:52
Maria MazurMaria Mazur
49.3k1360123
$endgroup$
$begingroup$
Sorry, I am terrible at maths. Can you please hint on how to proceed further?
$endgroup$
– user638500
Jan 29 at 14:07
add a comment |
$begingroup$
So $b=ax, c=ax^2$, and $d=ax^3$, and you have to prove: $$frac1{a^2x} + frac1{a^2x^5} > 2 (frac1{a^2x^4} + frac1{a^2x^2} - frac1{a^2x^3})$$
or $$1 + frac1{x^4} > 2 (frac1{x^3} + frac1{x} - frac1{x^2})$$
or $$boxed{x^4+1>2(x+x^3-x^2)}$$
or
$$x^4-2x^3+2x^2-2x+1>0iff (x^2-x)^2+(x-1)^2>0,$$ which is obviously true.
You can try to prove the boxed inequality also like this, using AM-GM inequality: $$x^4+x^2geq 2sqrt{x^4cdot x^2} = 2x^3$$ and$$x^2+1geq 2sqrt{x^2cdot 1} = 2x$$
so $$x^4+2x^2+x = (x^4+x^2)+(x^2+1)> 2x^3+2x$$ and we are done again.
edited Jan 29 at 14:45
J. W. Tanner
4,1961320
answered Jan 29 at 13:52
Maria MazurMaria Mazur
49.3k1360123
$endgroup$
So $b=ax, c=ax^2$, and $d=ax^3$, and you have to prove: $$frac1{a^2x} + frac1{a^2x^5} > 2 (frac1{a^2x^4} + frac1{a^2x^2} - frac1{a^2x^3})$$
or $$1 + frac1{x^4} > 2 (frac1{x^3} + frac1{x} - frac1{x^2})$$
or $$boxed{x^4+1>2(x+x^3-x^2)}$$
or
$$x^4-2x^3+2x^2-2x+1>0iff (x^2-x)^2+(x-1)^2>0,$$ which is obviously true.
You can try to prove the boxed inequality also like this, using AM-GM inequality: $$x^4+x^2geq 2sqrt{x^4cdot x^2} = 2x^3$$ and$$x^2+1geq 2sqrt{x^2cdot 1} = 2x$$
so $$x^4+2x^2+x = (x^4+x^2)+(x^2+1)> 2x^3+2x$$ and we are done again.
edited Jan 29 at 14:45
J. W. Tanner
4,1961320
answered Jan 29 at 13:52
Maria MazurMaria Mazur
49.3k1360123
edited Jan 29 at 14:45
J. W. Tanner
4,1961320
edited Jan 29 at 14:45
J. W. Tanner
4,1961320
edited Jan 29 at 14:45
J. W. Tanner
4,1961320
4,1961320
answered Jan 29 at 13:52
Maria MazurMaria Mazur
49.3k1360123
answered Jan 29 at 13:52
Maria MazurMaria Mazur
49.3k1360123
answered Jan 29 at 13:52
Maria MazurMaria Mazur
49.3k1360123
49.3k1360123
$begingroup$
Sorry, I am terrible at maths. Can you please hint on how to proceed further?
$endgroup$
– user638500
Jan 29 at 14:07
add a comment |
$begingroup$
Sorry, I am terrible at maths. Can you please hint on how to proceed further?
$endgroup$
– user638500
Jan 29 at 14:07
$begingroup$
Sorry, I am terrible at maths. Can you please hint on how to proceed further?
$endgroup$
– user638500
Jan 29 at 14:07
$begingroup$
Sorry, I am terrible at maths. Can you please hint on how to proceed further?
$endgroup$
– user638500
Jan 29 at 14:07
add a comment |
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