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Is it true that $sqrt(a_1+b_1+c_1)(a_2+b_2+c_2)geq sqrta_1a_2+sqrtb_1b_2+sqrtc_1c_2$? [closed]
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Let $a_1,b_1,c_1,a_2,b_2,c_2$ be positive real numbers. Is it true that
$sqrt(a_1+b_1+c_1)(a_2+b_2+c_2)geq sqrta_1a_2+sqrtb_1b_2+sqrtc_1c_2$
If this is true, then I will have proved an Olympiad problem. I couldn't find a counter-example.
inequality contest-math
asked Jul 18 at 1:43
fierydemon
4,32312152
closed as off-topic by Alex Francisco, Claude Leibovici, Gibbs, Jyrki Lahtonen, Adrian Keister Jul 18 at 13:34
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Claude Leibovici, Gibbs, Jyrki Lahtonen, Adrian Keister
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up vote
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Let $a_1,b_1,c_1,a_2,b_2,c_2$ be positive real numbers. Is it true that
$sqrt(a_1+b_1+c_1)(a_2+b_2+c_2)geq sqrta_1a_2+sqrtb_1b_2+sqrtc_1c_2$
If this is true, then I will have proved an Olympiad problem. I couldn't find a counter-example.
inequality contest-math
asked Jul 18 at 1:43
fierydemon
4,32312152
closed as off-topic by Alex Francisco, Claude Leibovici, Gibbs, Jyrki Lahtonen, Adrian Keister Jul 18 at 13:34
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Claude Leibovici, Gibbs, Jyrki Lahtonen, Adrian Keister
1
visual proof for this would have been great. could someone do that?
– Nick
Jul 18 at 3:13
1
@Nick the derivative proof below is in some sense "visual": you have that as you grow e.g. $a_1$, then the left-hand side grows as $fracddxsqrtx(a_2+b_2+c_2)$, while the right-hand side grows as $fracddxsqrtx(a_2)$, which is clearly less.
– Anonymous
Jul 18 at 23:35
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up vote
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up vote
4
down vote
favorite
Let $a_1,b_1,c_1,a_2,b_2,c_2$ be positive real numbers. Is it true that
$sqrt(a_1+b_1+c_1)(a_2+b_2+c_2)geq sqrta_1a_2+sqrtb_1b_2+sqrtc_1c_2$
If this is true, then I will have proved an Olympiad problem. I couldn't find a counter-example.
inequality contest-math
asked Jul 18 at 1:43
fierydemon
4,32312152
Let $a_1,b_1,c_1,a_2,b_2,c_2$ be positive real numbers. Is it true that
$sqrt(a_1+b_1+c_1)(a_2+b_2+c_2)geq sqrta_1a_2+sqrtb_1b_2+sqrtc_1c_2$
If this is true, then I will have proved an Olympiad problem. I couldn't find a counter-example.
inequality contest-math
asked Jul 18 at 1:43
fierydemon
4,32312152
asked Jul 18 at 1:43
fierydemon
4,32312152
asked Jul 18 at 1:43
fierydemon
4,32312152
asked Jul 18 at 1:43
fierydemon
4,32312152
4,32312152
closed as off-topic by Alex Francisco, Claude Leibovici, Gibbs, Jyrki Lahtonen, Adrian Keister Jul 18 at 13:34
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Claude Leibovici, Gibbs, Jyrki Lahtonen, Adrian Keister
closed as off-topic by Alex Francisco, Claude Leibovici, Gibbs, Jyrki Lahtonen, Adrian Keister Jul 18 at 13:34
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Claude Leibovici, Gibbs, Jyrki Lahtonen, Adrian Keister
1
visual proof for this would have been great. could someone do that?
– Nick
Jul 18 at 3:13
1
@Nick the derivative proof below is in some sense "visual": you have that as you grow e.g. $a_1$, then the left-hand side grows as $fracddxsqrtx(a_2+b_2+c_2)$, while the right-hand side grows as $fracddxsqrtx(a_2)$, which is clearly less.
– Anonymous
Jul 18 at 23:35
add a comment |Â
1
visual proof for this would have been great. could someone do that?
– Nick
Jul 18 at 3:13
1
@Nick the derivative proof below is in some sense "visual": you have that as you grow e.g. $a_1$, then the left-hand side grows as $fracddxsqrtx(a_2+b_2+c_2)$, while the right-hand side grows as $fracddxsqrtx(a_2)$, which is clearly less.
– Anonymous
Jul 18 at 23:35
1
1
visual proof for this would have been great. could someone do that?
– Nick
Jul 18 at 3:13
visual proof for this would have been great. could someone do that?
– Nick
Jul 18 at 3:13
1
1
@Nick the derivative proof below is in some sense "visual": you have that as you grow e.g. $a_1$, then the left-hand side grows as $fracddxsqrtx(a_2+b_2+c_2)$, while the right-hand side grows as $fracddxsqrtx(a_2)$, which is clearly less.
– Anonymous
Jul 18 at 23:35
@Nick the derivative proof below is in some sense "visual": you have that as you grow e.g. $a_1$, then the left-hand side grows as $fracddxsqrtx(a_2+b_2+c_2)$, while the right-hand side grows as $fracddxsqrtx(a_2)$, which is clearly less.
– Anonymous
Jul 18 at 23:35
add a comment |Â
2 Answers
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up vote
4
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Square both sides (they are non-negative) and rewrite the terms:
$$iff(sqrta_1sqrta_1+sqrtb_1sqrtb_1+sqrtc_1sqrtc_1)(sqrta_2sqrta_2+sqrtb_2sqrtb_2+sqrtc_2sqrtc_2)ge(sqrta_1sqrta_2+sqrtb_1sqrtb_2+sqrtc_1sqrtc_2)^2$$
This is precisely the Cauchy–Schwarz inequality applied on $(sqrta_1,sqrtb_1,sqrtc_1)$ and $(sqrta_2,sqrtb_2,sqrtc_2)$.
answered Jul 18 at 1:58
Parcly Taxel
33.6k136588
Thanks! Should have seen that! I was just scared about that inequality not being true I guess
– fierydemon
Jul 18 at 2:23
add a comment |Â
up vote
0
down vote
An alternative proof (to the "standard" Cauchy-Schwarz one) is to verify that the inequality holds (as an equality) when all the variables equal $0$; and that, when all variables are non-negative, the partial derivative of the left-hand side with respect to any variable equals or exceeds the partial derivative of the right-hand side (so, if you "grow" the variables from $0$ to whatever non-negative values you want, the "advantage" of the left-hand side can only grow).
In fact, by the same argument if all variables are strictly positive, the inequality becomes strict!
Note that this argument does require some care if you want to formalize it, because the derivative of $sqrtx$ with respect to $x$ diverges to $+infty$ as $x$ converges to $0$.
answered Jul 18 at 2:12
Anonymous
4,8033940
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Square both sides (they are non-negative) and rewrite the terms:
$$iff(sqrta_1sqrta_1+sqrtb_1sqrtb_1+sqrtc_1sqrtc_1)(sqrta_2sqrta_2+sqrtb_2sqrtb_2+sqrtc_2sqrtc_2)ge(sqrta_1sqrta_2+sqrtb_1sqrtb_2+sqrtc_1sqrtc_2)^2$$
This is precisely the Cauchy–Schwarz inequality applied on $(sqrta_1,sqrtb_1,sqrtc_1)$ and $(sqrta_2,sqrtb_2,sqrtc_2)$.
answered Jul 18 at 1:58
Parcly Taxel
33.6k136588
Thanks! Should have seen that! I was just scared about that inequality not being true I guess
– fierydemon
Jul 18 at 2:23
add a comment |Â
up vote
4
down vote
accepted
Square both sides (they are non-negative) and rewrite the terms:
$$iff(sqrta_1sqrta_1+sqrtb_1sqrtb_1+sqrtc_1sqrtc_1)(sqrta_2sqrta_2+sqrtb_2sqrtb_2+sqrtc_2sqrtc_2)ge(sqrta_1sqrta_2+sqrtb_1sqrtb_2+sqrtc_1sqrtc_2)^2$$
This is precisely the Cauchy–Schwarz inequality applied on $(sqrta_1,sqrtb_1,sqrtc_1)$ and $(sqrta_2,sqrtb_2,sqrtc_2)$.
answered Jul 18 at 1:58
Parcly Taxel
33.6k136588
Thanks! Should have seen that! I was just scared about that inequality not being true I guess
– fierydemon
Jul 18 at 2:23
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Square both sides (they are non-negative) and rewrite the terms:
$$iff(sqrta_1sqrta_1+sqrtb_1sqrtb_1+sqrtc_1sqrtc_1)(sqrta_2sqrta_2+sqrtb_2sqrtb_2+sqrtc_2sqrtc_2)ge(sqrta_1sqrta_2+sqrtb_1sqrtb_2+sqrtc_1sqrtc_2)^2$$
This is precisely the Cauchy–Schwarz inequality applied on $(sqrta_1,sqrtb_1,sqrtc_1)$ and $(sqrta_2,sqrtb_2,sqrtc_2)$.
answered Jul 18 at 1:58
Parcly Taxel
33.6k136588
Square both sides (they are non-negative) and rewrite the terms:
$$iff(sqrta_1sqrta_1+sqrtb_1sqrtb_1+sqrtc_1sqrtc_1)(sqrta_2sqrta_2+sqrtb_2sqrtb_2+sqrtc_2sqrtc_2)ge(sqrta_1sqrta_2+sqrtb_1sqrtb_2+sqrtc_1sqrtc_2)^2$$
This is precisely the Cauchy–Schwarz inequality applied on $(sqrta_1,sqrtb_1,sqrtc_1)$ and $(sqrta_2,sqrtb_2,sqrtc_2)$.
answered Jul 18 at 1:58
Parcly Taxel
33.6k136588
answered Jul 18 at 1:58
Parcly Taxel
33.6k136588
answered Jul 18 at 1:58
Parcly Taxel
33.6k136588
answered Jul 18 at 1:58
Parcly Taxel
33.6k136588
33.6k136588
Thanks! Should have seen that! I was just scared about that inequality not being true I guess
– fierydemon
Jul 18 at 2:23
add a comment |Â
Thanks! Should have seen that! I was just scared about that inequality not being true I guess
– fierydemon
Jul 18 at 2:23
Thanks! Should have seen that! I was just scared about that inequality not being true I guess
– fierydemon
Jul 18 at 2:23
Thanks! Should have seen that! I was just scared about that inequality not being true I guess
– fierydemon
Jul 18 at 2:23
add a comment |Â
up vote
0
down vote
An alternative proof (to the "standard" Cauchy-Schwarz one) is to verify that the inequality holds (as an equality) when all the variables equal $0$; and that, when all variables are non-negative, the partial derivative of the left-hand side with respect to any variable equals or exceeds the partial derivative of the right-hand side (so, if you "grow" the variables from $0$ to whatever non-negative values you want, the "advantage" of the left-hand side can only grow).
In fact, by the same argument if all variables are strictly positive, the inequality becomes strict!
Note that this argument does require some care if you want to formalize it, because the derivative of $sqrtx$ with respect to $x$ diverges to $+infty$ as $x$ converges to $0$.
answered Jul 18 at 2:12
Anonymous
4,8033940
add a comment |Â
up vote
0
down vote
An alternative proof (to the "standard" Cauchy-Schwarz one) is to verify that the inequality holds (as an equality) when all the variables equal $0$; and that, when all variables are non-negative, the partial derivative of the left-hand side with respect to any variable equals or exceeds the partial derivative of the right-hand side (so, if you "grow" the variables from $0$ to whatever non-negative values you want, the "advantage" of the left-hand side can only grow).
In fact, by the same argument if all variables are strictly positive, the inequality becomes strict!
Note that this argument does require some care if you want to formalize it, because the derivative of $sqrtx$ with respect to $x$ diverges to $+infty$ as $x$ converges to $0$.
answered Jul 18 at 2:12
Anonymous
4,8033940
add a comment |Â
up vote
0
down vote
up vote
0
down vote
An alternative proof (to the "standard" Cauchy-Schwarz one) is to verify that the inequality holds (as an equality) when all the variables equal $0$; and that, when all variables are non-negative, the partial derivative of the left-hand side with respect to any variable equals or exceeds the partial derivative of the right-hand side (so, if you "grow" the variables from $0$ to whatever non-negative values you want, the "advantage" of the left-hand side can only grow).
In fact, by the same argument if all variables are strictly positive, the inequality becomes strict!
Note that this argument does require some care if you want to formalize it, because the derivative of $sqrtx$ with respect to $x$ diverges to $+infty$ as $x$ converges to $0$.
answered Jul 18 at 2:12
Anonymous
4,8033940
An alternative proof (to the "standard" Cauchy-Schwarz one) is to verify that the inequality holds (as an equality) when all the variables equal $0$; and that, when all variables are non-negative, the partial derivative of the left-hand side with respect to any variable equals or exceeds the partial derivative of the right-hand side (so, if you "grow" the variables from $0$ to whatever non-negative values you want, the "advantage" of the left-hand side can only grow).
In fact, by the same argument if all variables are strictly positive, the inequality becomes strict!
Note that this argument does require some care if you want to formalize it, because the derivative of $sqrtx$ with respect to $x$ diverges to $+infty$ as $x$ converges to $0$.
answered Jul 18 at 2:12
Anonymous
4,8033940
edited Jul 18 at 2:17
answered Jul 18 at 2:12
Anonymous
4,8033940
answered Jul 18 at 2:12
Anonymous
4,8033940
answered Jul 18 at 2:12
Anonymous
4,8033940
4,8033940
add a comment |Â
add a comment |Â
1
visual proof for this would have been great. could someone do that?
– Nick
Jul 18 at 3:13
1
@Nick the derivative proof below is in some sense "visual": you have that as you grow e.g. $a_1$, then the left-hand side grows as $fracddxsqrtx(a_2+b_2+c_2)$, while the right-hand side grows as $fracddxsqrtx(a_2)$, which is clearly less.
– Anonymous
Jul 18 at 23:35