Prove: "a√(b/c)+b√(c/a)+c√(a/b)≥1

[Andrax]

Homework Statement

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

The Attempt at a Solution

i tried the following
x$\sqrt{abc}$ : ab$\sqrt{a}$+bc$\sqrt{b}$+ac$\sqrt{c}$$\geq$ $\sqrt{abc}$
Also saying that a bc from ]0,+infini[ and a+b+c=1 means that a and b and c $\leq$ 1
i tried the inequality forgot it's name $\sqrt{abc}$+1$\geq$$\sqrt{abc}$*(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 .

 

[Joffan]

Homework Statement

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

The Attempt at a Solution

i tried the following
x$\sqrt{abc}$ : ab$\sqrt{a}$+bc$\sqrt{b}$+ac$\sqrt{c}$$\geq$ $\sqrt{abc}$
Also saying that a bc from ]0,+infini[ and a+b+c=1 means that a and b and c $\leq$ 1
i tried the inequality forgot it's name $\sqrt{abc}$+1$\geq$$\sqrt{abc}$*(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. $(\sqrt a + \sqrt b + \sqrt c)^2$?

 

[Andrax]

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)

 

[Andrax]

didn't work :/ I'm stuck lol

 

[haruspex]

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.

 

[Ray Vickson]

Homework Statement

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

The Attempt at a Solution

i tried the following
x$\sqrt{abc}$ : ab$\sqrt{a}$+bc$\sqrt{b}$+ac$\sqrt{c}$$\geq$ $\sqrt{abc}$
Also saying that a bc from ]0,+infini[ and a+b+c=1 means that a and b and c $\leq$ 1
i tried the inequality forgot it's name $\sqrt{abc}$+1$\geq$$\sqrt{abc}$*(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
$$f(a,b,c) = a \sqrt{\frac{b}{c}} + b \sqrt{\frac{c}{a}} + c \sqrt{\frac{a}{b}}$$ and $g(a,b,c) = a+b+c$, we can look at the problem
$$\min f(a,b,c)\\ \text{subject to } \\ g(a,b,c) = 1\\ a \geq b \\ b \geq c .$$
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
$$F_a = \frac{N}{D}, \: N = 2 a^2 b -\sqrt{a}b^{3/2}+a^{3/2}, \; D = 2 a^2 \sqrt{b},$$
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.

 

[Andrax]

This problem is solvable via constrained optimization. If
$$f(a,b,c) = a \sqrt{\frac{b}{c}} + b \sqrt{\frac{c}{a}} + c \sqrt{\frac{a}{b}}$$ and $g(a,b,c) = a+b+c$, we can look at the problem
$$\min f(a,b,c)\\ \text{subject to } \\ g(a,b,c) = 1\\ a \geq b \\ b \geq c .$$
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
$$F_a = \frac{N}{D}, \: N = 2 a^2 b -\sqrt{a}b^{3/2}+a^{3/2}, \; D = 2 a^2 \sqrt{b},$$
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

 

[Joffan]

Ray's is an impressive approach, but I'm not convinced about this step:

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.

 

[Ray Vickson]

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
$$\min f(a,b,c) = a b^{1/2}c^{-1/2} + b c^{1/2}a^{-1/2} + c a^{1/2} b^{-1/2} \\ \text{subject to}\\ g_1(a,b,c) \equiv a^{-1} b \leq 1\\ c \geq 1$$
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