\[ \large (ab+bc+ca) \left (\frac 1{(a+b)^2} + \frac 1{(b+c)^2}+\frac 1{(c+a)^2} \right) \]

For \(a, b, c\) are positive reals, if the infinium (minimum) value of the expression above can be expressed as \( \frac{ x}{y} \) for positive coprime integers \(x\) and \(y\), find the value of \( x - y \).

This problem is already posted but it has no solutions and I wanted to know it and so I am posting it again.

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## Comments

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TopNewestWhy don't you link us to the problem itself?

What have you tried? Have you tried homogenizing \(a,b,c\)? That is, set one of the symmetric sums of \(a,b,c\) to be a certain constant.

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Sir, I don't know how to create a link. Can you please explain me what are you trying to convey through 'homogenizing \(a, b, c\)'?

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I've fixed the link for you.

Synopsis of homogenizing the variables: If the minimum value occurs when \((a,b,c) = (a', b' , c') \), then the minimum value also occurs when \((a,b,c) = (ka', kb', kc') \), where \(k \) is any positive number.

So in this case, you can safely make the assumption that \(a + b = c = L \), \(ab + ac + bc = L \) or \(abc = L \) for some positive constant \(L \).

Now suppose I assumed that \(a+b+c = L \), can you simplify the desired expression in terms of \(L \)? What classical inequalities can we use? Maybe Cauchy-Schwarz inequality? Maybe Titu's lemma?

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Sir , I tried doing it but couldn't crack it.

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As for the inequalities , I know almost all of them.

I tried solving it directly using Cauchy-Schwarz inequality.

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Then apply AM-HM for the numbers (1/(a+b) + 1/(b + c) + 1/(a+c))

WLOG, set a+b+c = 1

What's left is to minimize (ab+ac+bc) subject to a,b,c>0 and a+b+c=1.

Use Lagrange multipliers to finish it off, and you will get min(ab+ac+bc) occurs when a=b=c

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If yes, I'll just post the calculus approach here and you can submit the full solution in the problem itself. How about that?

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We want to minimize \(ab + bc + ac\) subject to the constraints \(a,b,c> 0 \) and \(a+b+c= 1\).

Substituting \(c = 1 -a-b\) into this given expression gives \(a + b - a^2 - b^2 - ab \).

Let \(f = a+b - a^2 - b^2 - ab \). Since both \(a,b\) are in the open interval \((0,1) \), then \(f \) has no boundary points. So the min/max of \(f\) must occur at its extremal point(s).

Taking the partial derivative of \(f\) with respect to \(a\), we get \(f_a = 1 - 2a - b \).

Similarly, \(f_b = 1 - 2b - a \).

At the extremal point(s), \(f_a = f_b = 0 \). Solving this gives \(a = b= \dfrac13\Rightarrow f = \dfrac13\).

Now, we want to prove that \(f = \dfrac13\) is a minimum value. To do so, we apply the second derivative test determinant:

\[ D = f_{aa} f_{bb} - (f_{ab})^2 , \]

where \(f_{aa} = \dfrac{\partial^2}{\partial a^2} f = -2 \), \(f_{bb} = \dfrac{\partial^2}{\partial b^2} f = -2 \) and \(f_{ab} = \dfrac{\partial^2}{\partial a \partial b} f = -1 \).

This gives \(D = (-2)(-2) - (-1)^2 = 3 > 0 \) with \(f_{aa} \left ( \dfrac13, \dfrac13\right) = \dfrac13 > 0 \), thus the point we have found is indeed a minimum point.

In other words, we have shown that the expression \(ab+bc+ac\) is minimized when \(a =b=c= \dfrac13\) (under those stated constraints).

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In the Weighted AM -GM Inequality , are the weights positive reals , fractions or positive integers ?

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My bad, the answer should be "positive integers".

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Anyway, the reason why the weights can be non-integers as well is because AMGM is a derivation from the Jensen's inequality. And because Jensen's inequality doesn't specify that the weights need to be integers, then neither does AMGM as well.

Note that a^3.5 cannot be expressed as (a * a * a ... * a) (3.5 times), same goes for the AMGM stuffs that you've mentioned.

On the other hand, I've found a (correct) way of proving that min(ab + ac + bc) occurs when a=b=c.

Hint: Start with (a+b+c)^2 =a^2+b^2+c^2 + 2(ab+ac+bc), so we want to maximize (a^2 + b^2 + c^2) subject to the same constraints. You can either solve by Lagrange multipliers (followed by Hessian matrix) or interpret the expression (a^2+b^2+c^2) as an equation of a sphere and the equation (a+b+c=1) as an equation of a plane in the first quadrant.

I'm actually still not satisfied with my approach because it uses calculus approach, which is a little bit unacceptable in my opinion. I'll try to post a full inequalities approach (and no calculus method) if I'm able to.

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On the other hand, you can't guarantee that if you raised it "n number of times" then it will be an integer. You could have weights of irrational numbers like (sqrt2) as well, so no matter how what integer value of "n' you raise, you can't get an integer.

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\(a_1 + a_2 + a_3 + \cdot \cdot \cdot + a_n \geq n\cdot {(a_1a_2a_3\cdot \cdot \cdot a_n)}^{\frac{1}{n}}\)

Jensen' s Inequality was used as :

Take \(f(x)\) as \(ln(x)\) , then since \(f"(x) < 0\).

\(\Rightarrow \frac{f(a_1) + f(a_2) + f(a_3) + \cdot \cdot \cdot + f(a_n)}{n} \leq f(\frac{a_1 +a_2 +a_3 +\cdot \cdot \cdot + a_n}{n})\).

So to get the weighted result we'll have to do this (Suppose for two terms):

\(\frac{\underbrace{f(a_1) + f(a_1) + f(a_1) + \cdot \cdot \cdot + f(a_1)}_{\text{b_1 times}} + \underbrace{f(a_2) + f(a_2) + f(a_2) + \cdot \cdot \cdot + f(a_2)}_{\text{b_2 times}}}{b_1 + b_2} \leq f(\frac{\underbrace{a_1 +a_1 + \cdot \cdot \cdot + a_1}_{\text{b_1 times}} + \underbrace{a_2 + a_2 + \cdot \cdot \cdot + a_2}_{\text{b_2 times}}}{b_1 +b_2})\)

So , won't this force \(b_i\) to be a positive integer?

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Here's the link

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An easy one ans is 5

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Please post your complete solution in that case.

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Ya sure.. but currently my pc is not working. As soon as I get it back most probably by 3 days... I will post

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