Prove: "a√(b/c)+b√(c/a)+c√(a/b)≥1

  • Thread starter Andrax
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In summary: N_1## in region X is > 0, so ##F_a > 0## in X. That is, F(a,b) is strictly increasing in a, for a ≥ b. Therefore, the constrained minimum of F must occur along the line a = b, so it is enough to look at ##F(b,b) = 1 + \sqrt{b} + b^{3/2},## which we want to minimize for b ≥ 1. The optimal solution is at (a,b) = (1,1), hence we have the optimum (a,b,c) = (1,1,1). Now we just need to re-scale to (a,b,c) =
  • #1
Andrax
117
0

Homework Statement


let a b c real numbers [itex]\in[/itex] ]0,+infini[
we have a+b+c=1 and a [itex]\geq[/itex] b [itex]\geq[/itex] c
Prove that a[itex]\sqrt{\frac{b}{c}}[/itex]+b[itex]\sqrt{\frac{c}{a}}[/itex]+c[itex]\sqrt{\frac{a}{b}}[/itex] [itex]\geq[/itex] 1

The Attempt at a Solution


i tried the following
x[itex]\sqrt{abc}[/itex] : ab[itex]\sqrt{a}[/itex]+bc[itex]\sqrt{b}[/itex]+ac[itex]\sqrt{c}[/itex][itex]\geq[/itex] [itex]\sqrt{abc}[/itex]
Also saying that a bc from ]0,+infini[ and a+b+c=1 means that a and b and c [itex]\leq[/itex] 1
i tried the inequality forgot it's name [itex]\sqrt{abc}[/itex]+1[itex]\geq[/itex][itex]\sqrt{abc}[/itex]*(a+b+c) but it dosen't give the wanted result well I've been working on this exercises for 4 days in a row this is an olympiad exercise btw so yeah help will be appreciated .
 
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  • #2
Andrax said:

Homework Statement


let a b c real numbers [itex]\in[/itex] ]0,+infini[
we have a+b+c=1 and a [itex]\geq[/itex] b [itex]\geq[/itex] c
Prove that a[itex]\sqrt{\frac{b}{c}}[/itex]+b[itex]\sqrt{\frac{c}{a}}[/itex]+c[itex]\sqrt{\frac{a}{c}}[/itex] [itex]\geq[/itex] 1



The Attempt at a Solution


i tried the following
x[itex]\sqrt{abc}[/itex] : ab[itex]\sqrt{a}[/itex]+bc[itex]\sqrt{b}[/itex]+ac[itex]\sqrt{c}[/itex][itex]\geq[/itex] [itex]\sqrt{abc}[/itex]
Also saying that a bc from ]0,+infini[ and a+b+c=1 means that a and b and c [itex]\leq[/itex] 1
i tried the inequality forgot it's name [itex]\sqrt{abc}[/itex]+1[itex]\geq[/itex][itex]\sqrt{abc}[/itex]*(a+b+c) but it dosen't give the wanted result well I've been working on this exercises for 4 days in a row this is an olympiad exercise btw so yeah help will be appreciated .

I assume you've got a typo there and you're meant to prove that ## a \sqrt{\frac{b}{c}}+b \sqrt{\frac{c}{a}}+c \sqrt{\frac{a}{b}} \geq 1 ##

And 1 is not a possible value for a,b,c; we can say ##1 \gt a\geq b\geq c\gt 0##

I don't know the answer, but have you tried starting from somewhere else, eg. ##(\sqrt a + \sqrt b + \sqrt c)^2##?
 
  • #3
Joffan said:
I assume you've got a typo there and you're meant to prove that ## a \sqrt{\frac{b}{c}}+b \sqrt{\frac{c}{a}}+c \sqrt{\frac{a}{b}} \geq 1 ##

And 1 is not a possible value for a,b,c; we can say ##1 \gt a\geq b\geq c\gt 0##

I don't know the answer, but have you tried starting from somewhere else, eg. ##(\sqrt a + \sqrt b + \sqrt c)^2##?
yes i had a typo thanks , hm intersting i'll try that thanks (probably won't work anyway)
 
  • #4
didn't work :/ I'm stuck lol
 
  • #5
I think I managed to get it into the form "show c2(bc-a2)+a2(ca-b2)+b2(ab-c2) >= 0 for all positive a, b, c". Looks better, but don't see where to go from there.
 
  • #6
Andrax said:

Homework Statement


let a b c real numbers [itex]\in[/itex] ]0,+infini[
we have a+b+c=1 and a [itex]\geq[/itex] b [itex]\geq[/itex] c
Prove that a[itex]\sqrt{\frac{b}{c}}[/itex]+b[itex]\sqrt{\frac{c}{a}}[/itex]+c[itex]\sqrt{\frac{a}{b}}[/itex] [itex]\geq[/itex] 1



The Attempt at a Solution


i tried the following
x[itex]\sqrt{abc}[/itex] : ab[itex]\sqrt{a}[/itex]+bc[itex]\sqrt{b}[/itex]+ac[itex]\sqrt{c}[/itex][itex]\geq[/itex] [itex]\sqrt{abc}[/itex]
Also saying that a bc from ]0,+infini[ and a+b+c=1 means that a and b and c [itex]\leq[/itex] 1
i tried the inequality forgot it's name [itex]\sqrt{abc}[/itex]+1[itex]\geq[/itex][itex]\sqrt{abc}[/itex]*(a+b+c) but it dosen't give the wanted result well I've been working on this exercises for 4 days in a row this is an olympiad exercise btw so yeah help will be appreciated .

This problem is solvable via constrained optimization. If
[tex] f(a,b,c) = a \sqrt{\frac{b}{c}} + b \sqrt{\frac{c}{a}} + c \sqrt{\frac{a}{b}}[/tex] and ## g(a,b,c) = a+b+c##, we can look at the problem
[tex] \min f(a,b,c)\\
\text{subject to } \\
g(a,b,c) = 1\\
a \geq b \\
b \geq c . [/tex]
If the minimal value of f is ≥ 1 we are done. Note that f , g and all the inequality constraint functions are homogeneous of degree 1, so if (a,b,c) solves the problem with g = m, then (a/m,b/m,c/m) solves it with g = 1. Therefore, we might as well fix c = 1, then solve the problem of minimizing F(a,b) = f(a,b,1) subject to the conditions a ≥ b ≥ 1. We can show that the point (a,b) = (1,1) satisfies the Karush-Kuhn-Tucker conditions for this problem so is a (local) constrained min. To show it is the global min requires a bit more work.

Let ##F_a = \partial F / \partial a##, and note that
[tex] F_a = \frac{N}{D}, \: N = 2 a^2 b -\sqrt{a}b^{3/2}+a^{3/2}, \; D = 2 a^2 \sqrt{b},[/tex]
so ##\text{sign}(F_a) = \text{sign}(N)##. We want to show that N > 0 on ##X = \{a \geq b \geq 1\},## so look at ##N_a = \partial N /\partial a = N_1 /D_1,## where ##D_1 = 2 \sqrt{a}## and ##N_1 = 3a + 8 a^{3/2} b - b^{3/2}.## By taking a derivative again it is not hard to show that the minimum of ##N_1## in region X is > 0, so ##F_a > 0## in X. That is, F(a,b) is strictly increasing in a, for a ≥ b. Therefore, the constrained minimum of F must occur along the line a = b, so it is enough to look at ##F(b,b) = 1 + \sqrt{b} + b^{3/2},## which we want to minimize for b ≥ 1. The optimal solution is at (a,b) = (1,1), hence we have the optimum (a,b,c) = (1,1,1). Now we just need to re-scale to (a,b,c) = (1/3,1/3,1/3) to get the minimum of f for g = 1.
 
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  • #7
Ray Vickson said:
This problem is solvable via constrained optimization. If
[tex] f(a,b,c) = a \sqrt{\frac{b}{c}} + b \sqrt{\frac{c}{a}} + c \sqrt{\frac{a}{b}}[/tex] and ## g(a,b,c) = a+b+c##, we can look at the problem
[tex] \min f(a,b,c)\\
\text{subject to } \\
g(a,b,c) = 1\\
a \geq b \\
b \geq c . [/tex]
If the minimal value of f is ≥ 1 we are done. Note that f , g and all the inequality constraint functions are homogeneous of degree 1, so if (a,b,c) solves the problem with g = m, then (a/m,b/m,c/m) solves it with g = 1. Therefore, we might as well fix c = 1, then solve the problem of minimizing F(a,b) = f(a,b,1) subject to the conditions a ≥ b ≥ 1. We can show that the point (a,b) = (1,1) satisfies the Karush-Kuhn-Tucker conditions for this problem so is a (local) constrained min. To show it is the global min requires a bit more work.

Let ##F_a = \partial F / \partial a##, and note that
[tex] F_a = \frac{N}{D}, \: N = 2 a^2 b -\sqrt{a}b^{3/2}+a^{3/2}, \; D = 2 a^2 \sqrt{b},[/tex]
so ##\text{sign}(F_a) = \text{sign}(N)##. We want to show that N > 0 on ##X = \{a \geq b \geq 1\},## so look at ##N_a = \partial N /\partial a = N_1 /D_1,## where ##D_1 = 2 \sqrt{a}## and ##N_1 = 3a + 8 a^{3/2} b - b^{3/2}.## By taking a derivative again it is not hard to show that the minimum of ##N_1## in region X is > 0, so ##F_a > 0## in X. That is, F(a,b) is strictly increasing in a, for a ≥ b. Therefore, the constrained minimum of F must occur along the line a = b, so it is enough to look at ##F(b,b) = 1 + \sqrt{b} + b^{3/2},## which we want to minimize for b ≥ 1. The optimal solution is at (a,b) = (1,1), hence we have the optimum (a,b,c) = (1,1,1). Now we just need to re-scale to (a,b,c) = (1/3,1/3,1/3) to get the minimum of f for g = 1.

thank you
 
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  • #8
Ray's is an impressive approach, but I'm not convinced about this step:
Ray Vickson said:
Note that f , g and all the inequality constraint functions are homogeneous of degree 1, so if (a,b,c) solves the problem with g = m, then (a/m,b/m,c/m) solves it with g = 1. Therefore, we might as well fix c = 1, then solve the problem of minimizing F(a,b) = f(a,b,1) subject to the conditions a ≥ b ≥ 1.
It seems to me that we are not minimizing for a particular m - rather we are allowing m to vary. The fact that a and b end up at their constraints does nothing to dispel this uneasiness.
 
  • #9
Joffan said:
Ray's is an impressive approach, but I'm not convinced about this step:

It seems to me that we are not minimizing for a particular m - rather we are allowing m to vary. The fact that a and b end up at their constraints does nothing to dispel this uneasiness.

Neglecting the sum constraint is OK (see below). Perhaps a more damaging objection is that by first setting c = 1, then solving the (a,b) problem (without a sum constraint) we can lose the optimum. That does not happen, but it is probably better to work with the full 3-variable problem. We can solve for (a,b,c) without a sum constraint by recognizing that
[tex] \min f(a,b,c) = a b^{1/2}c^{-1/2} + b c^{1/2}a^{-1/2} + c a^{1/2} b^{-1/2} \\
\text{subject to}\\
g_1(a,b,c) \equiv a^{-1} b \leq 1\\
c \geq 1
[/tex]
is a posynomial geometric programming problem--that is, f and g_1 posynomials--so any Kuhn-Tucker point is a global optimum, despite the lack of convexity. (Essentially, this follows by writing a = exp(u), b = exp(v) and c = exp(w), and noting that we have a convex programming problem in the variables u, v and w.) It is easy to verify that (a,b,c) = (1,1,1) is a Kuhn-Tucker point. In addition it satisfies the missing constraint b >= c, hence is *also a global solution when that constraint is added to the problem*.

Now: about the sum constraint. Without that constraint we get a solution that satisfies sum = 3, so it is also the optimal solution of the problem when the constraint sum = 3 is added to the problem. In other words we have f(1,1,1) < f(a,b,c) for all (a,b,c) that satisfy the two inequality constraints and the sum constraint a+b+c = 3. Thus, f(1/3,1/3,1/3)= (1/3)f(1,1,1) < (1/3)f(a,b,c) = f(a',b',c') for any (a',b',c') = (a/3,b/3,c/3) that satisfy the constraint a'+b'+c'=1,

Finally, for definitions and properties of posynomials, see, eg.,
http://en.wikipedia.org/wiki/Posynomial
or
https://inst.eecs.berkeley.edu/~ee127a/book/login/l_gp_posy.html
 
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FAQ: Prove: "a√(b/c)+b√(c/a)+c√(a/b)≥1

What is the meaning of the expression "a√(b/c)+b√(c/a)+c√(a/b)"?

The expression "a√(b/c)+b√(c/a)+c√(a/b)" is an inequality that represents a mathematical relationship between three variables, a, b, and c. The symbol √ represents the square root function, and the division symbol / indicates division. The inequality states that the sum of the square roots of a times the square root of b divided by c, b times the square root of c divided by a, and c times the square root of a divided by b is greater than or equal to 1.

What does it mean to "prove" this inequality?

To prove this inequality means to demonstrate that it is always true for any values of a, b, and c that satisfy certain conditions. This involves using mathematical reasoning and techniques to show that the inequality holds true, regardless of the specific values of the variables.

What are the conditions that need to be satisfied for this inequality to hold true?

In order for this inequality to hold true, the variables a, b, and c must be positive real numbers. Additionally, at least one of the variables must be greater than or equal to 1, and the sum of the three variables must be greater than or equal to 3.

Can this inequality be proven using algebra?

Yes, this inequality can be proven using algebraic manipulation and properties of inequalities. By rearranging the terms and applying rules of inequality, it is possible to arrive at a statement that is always true, thus proving the original inequality.

What are some real-life applications of this inequality?

This inequality can be applied in various fields such as economics, physics, and engineering. For example, in economics, it can be used to prove the relationship between costs and production in a business. In physics, it can be used to represent the relationship between different physical quantities. In engineering, it can be used to optimize the use of resources in a project.

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