How to show that $k_1k_2 - k_0 > 0$?
$begingroup$
Let $a_1, a_2, b, c in mathbb{R}$ such that
$ a_1 neq 0,, $
$a_2 neq 0,, $
$ a_1 > a_2,, $
$ b neq 0,,$
$c neq 0,,$
and $ b_1 < b < b_2 < 2(a_1 - a_2)$,
where :
$$b_1 = frac{2(a_1-a_2)}{sqrt{1+frac{4a_1}{c}} + 1} qquad b_2 = frac{2(a_1-a_2)}{sqrt{1+frac{4a_2}{c}} + 1}$$
Prove that $k_1k_2 - k_0 > 0,$
where :
$$k = frac{a_1a_2c}{b^3(a_1-a_2)}left(b+frac{2(a_1-a_2)}{sqrt{1+frac{4a_1}{c}} - 1}right)left(frac{2(a_1-a_2)}{sqrt{1+frac{4a_1}{c}} + 1}right)$$
$$k_2 = frac{(a_1+a_2)b^2+2c(a_1-a_2)^2}{(a_1-a_2)b}$$
$$k_1 = cleft[frac2b -frac{1}{a_1 - a_2}right][(a_1 + a_2)b + (a_1-a_2)c]$$
$$k_0 = -k(b-b_1)(b-b_2)$$
This was a Bonus question in a test I took last week, I computed $k_1k_2$ and kept doing circular simplifications but I never reached something that looks like $k_0$.
Is there some kind of observation to make to turn this into a less painful problem ?
inequality
edited Jan 11 at 13:47
Did
249k23228466
asked Jan 10 at 15:14
rapidracimrapidracim
1,7441419
$endgroup$
add a comment |
$begingroup$
Let $a_1, a_2, b, c in mathbb{R}$ such that
$ a_1 neq 0,, $
$a_2 neq 0,, $
$ a_1 > a_2,, $
$ b neq 0,,$
$c neq 0,,$
and $ b_1 < b < b_2 < 2(a_1 - a_2)$,
where :
$$b_1 = frac{2(a_1-a_2)}{sqrt{1+frac{4a_1}{c}} + 1} qquad b_2 = frac{2(a_1-a_2)}{sqrt{1+frac{4a_2}{c}} + 1}$$
Prove that $k_1k_2 - k_0 > 0,$
where :
$$k = frac{a_1a_2c}{b^3(a_1-a_2)}left(b+frac{2(a_1-a_2)}{sqrt{1+frac{4a_1}{c}} - 1}right)left(frac{2(a_1-a_2)}{sqrt{1+frac{4a_1}{c}} + 1}right)$$
$$k_2 = frac{(a_1+a_2)b^2+2c(a_1-a_2)^2}{(a_1-a_2)b}$$
$$k_1 = cleft[frac2b -frac{1}{a_1 - a_2}right][(a_1 + a_2)b + (a_1-a_2)c]$$
$$k_0 = -k(b-b_1)(b-b_2)$$
This was a Bonus question in a test I took last week, I computed $k_1k_2$ and kept doing circular simplifications but I never reached something that looks like $k_0$.
Is there some kind of observation to make to turn this into a less painful problem ?
inequality
edited Jan 11 at 13:47
Did
249k23228466
asked Jan 10 at 15:14
rapidracimrapidracim
1,7441419
$endgroup$
$begingroup$
"this was a Bonus question in a test I took last week" I am curious, in which (kind of) curriculum?
$endgroup$
– Did
Jan 11 at 9:19
$begingroup$
@Did Advanced ODE's, there is many problems with this same concept, you can use the results to show positive definiteness of some complicated lyapunov functions
$endgroup$
– rapidracim
Jan 11 at 9:49
$begingroup$
This might be homogeneous in $c$ hence solving the case $c=1$ would suffice. Apart from that...
$endgroup$
– Did
Jan 11 at 13:57
add a comment |
$begingroup$
Let $a_1, a_2, b, c in mathbb{R}$ such that
$ a_1 neq 0,, $
$a_2 neq 0,, $
$ a_1 > a_2,, $
$ b neq 0,,$
$c neq 0,,$
and $ b_1 < b < b_2 < 2(a_1 - a_2)$,
where :
$$b_1 = frac{2(a_1-a_2)}{sqrt{1+frac{4a_1}{c}} + 1} qquad b_2 = frac{2(a_1-a_2)}{sqrt{1+frac{4a_2}{c}} + 1}$$
Prove that $k_1k_2 - k_0 > 0,$
where :
$$k = frac{a_1a_2c}{b^3(a_1-a_2)}left(b+frac{2(a_1-a_2)}{sqrt{1+frac{4a_1}{c}} - 1}right)left(frac{2(a_1-a_2)}{sqrt{1+frac{4a_1}{c}} + 1}right)$$
$$k_2 = frac{(a_1+a_2)b^2+2c(a_1-a_2)^2}{(a_1-a_2)b}$$
$$k_1 = cleft[frac2b -frac{1}{a_1 - a_2}right][(a_1 + a_2)b + (a_1-a_2)c]$$
$$k_0 = -k(b-b_1)(b-b_2)$$
This was a Bonus question in a test I took last week, I computed $k_1k_2$ and kept doing circular simplifications but I never reached something that looks like $k_0$.
Is there some kind of observation to make to turn this into a less painful problem ?
inequality
edited Jan 11 at 13:47
Did
249k23228466
asked Jan 10 at 15:14
rapidracimrapidracim
1,7441419
$endgroup$
Let $a_1, a_2, b, c in mathbb{R}$ such that
$ a_1 neq 0,, $
$a_2 neq 0,, $
$ a_1 > a_2,, $
$ b neq 0,,$
$c neq 0,,$
and $ b_1 < b < b_2 < 2(a_1 - a_2)$,
where :
$$b_1 = frac{2(a_1-a_2)}{sqrt{1+frac{4a_1}{c}} + 1} qquad b_2 = frac{2(a_1-a_2)}{sqrt{1+frac{4a_2}{c}} + 1}$$
Prove that $k_1k_2 - k_0 > 0,$
where :
$$k = frac{a_1a_2c}{b^3(a_1-a_2)}left(b+frac{2(a_1-a_2)}{sqrt{1+frac{4a_1}{c}} - 1}right)left(frac{2(a_1-a_2)}{sqrt{1+frac{4a_1}{c}} + 1}right)$$
$$k_2 = frac{(a_1+a_2)b^2+2c(a_1-a_2)^2}{(a_1-a_2)b}$$
$$k_1 = cleft[frac2b -frac{1}{a_1 - a_2}right][(a_1 + a_2)b + (a_1-a_2)c]$$
$$k_0 = -k(b-b_1)(b-b_2)$$
This was a Bonus question in a test I took last week, I computed $k_1k_2$ and kept doing circular simplifications but I never reached something that looks like $k_0$.
Is there some kind of observation to make to turn this into a less painful problem ?
inequality
inequality
edited Jan 11 at 13:47
Did
249k23228466
asked Jan 10 at 15:14
rapidracimrapidracim
1,7441419
edited Jan 11 at 13:47
Did
249k23228466
asked Jan 10 at 15:14
rapidracimrapidracim
1,7441419
edited Jan 11 at 13:47
Did
249k23228466
edited Jan 11 at 13:47
Did
249k23228466
edited Jan 11 at 13:47
Did
249k23228466
249k23228466
asked Jan 10 at 15:14
rapidracimrapidracim
1,7441419
asked Jan 10 at 15:14
rapidracimrapidracim
1,7441419
asked Jan 10 at 15:14
rapidracimrapidracim
1,7441419
1,7441419
$begingroup$
"this was a Bonus question in a test I took last week" I am curious, in which (kind of) curriculum?
$endgroup$
– Did
Jan 11 at 9:19
$begingroup$
@Did Advanced ODE's, there is many problems with this same concept, you can use the results to show positive definiteness of some complicated lyapunov functions
$endgroup$
– rapidracim
Jan 11 at 9:49
$begingroup$
This might be homogeneous in $c$ hence solving the case $c=1$ would suffice. Apart from that...
$endgroup$
– Did
Jan 11 at 13:57
add a comment |
$begingroup$
"this was a Bonus question in a test I took last week" I am curious, in which (kind of) curriculum?
$endgroup$
– Did
Jan 11 at 9:19
$begingroup$
@Did Advanced ODE's, there is many problems with this same concept, you can use the results to show positive definiteness of some complicated lyapunov functions
$endgroup$
– rapidracim
Jan 11 at 9:49
$begingroup$
This might be homogeneous in $c$ hence solving the case $c=1$ would suffice. Apart from that...
$endgroup$
– Did
Jan 11 at 13:57
$begingroup$
"this was a Bonus question in a test I took last week" I am curious, in which (kind of) curriculum?
$endgroup$
– Did
Jan 11 at 9:19
$begingroup$
"this was a Bonus question in a test I took last week" I am curious, in which (kind of) curriculum?
$endgroup$
– Did
Jan 11 at 9:19
$begingroup$
@Did Advanced ODE's, there is many problems with this same concept, you can use the results to show positive definiteness of some complicated lyapunov functions
$endgroup$
– rapidracim
Jan 11 at 9:49
$begingroup$
@Did Advanced ODE's, there is many problems with this same concept, you can use the results to show positive definiteness of some complicated lyapunov functions
$endgroup$
– rapidracim
Jan 11 at 9:49
$begingroup$
This might be homogeneous in $c$ hence solving the case $c=1$ would suffice. Apart from that...
$endgroup$
– Did
Jan 11 at 13:57
$begingroup$
This might be homogeneous in $c$ hence solving the case $c=1$ would suffice. Apart from that...
$endgroup$
– Did
Jan 11 at 13:57
add a comment |
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$begingroup$
"this was a Bonus question in a test I took last week" I am curious, in which (kind of) curriculum?
$endgroup$
– Did
Jan 11 at 9:19
$begingroup$
@Did Advanced ODE's, there is many problems with this same concept, you can use the results to show positive definiteness of some complicated lyapunov functions
$endgroup$
– rapidracim
Jan 11 at 9:49
$begingroup$
This might be homogeneous in $c$ hence solving the case $c=1$ would suffice. Apart from that...
$endgroup$
– Did
Jan 11 at 13:57