# Inequality

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

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.

Note by Ankit Kumar Jain
3 years, 7 months ago

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An easy one ans is 5

- 3 years, 7 months ago

- 3 years, 7 months ago

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

- 3 years, 7 months ago

Okay! :)

- 3 years, 7 months ago

- 3 years, 7 months ago

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

- 3 years, 7 months ago

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?

- 3 years, 7 months ago

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

- 3 years, 7 months ago

What inequalities have you tried? What haven't you tried? Are you familiar with all (or most) the common classical inequalities?

- 3 years, 7 months ago

Sir , I couldn't simplify the expressions in terms of L to make it something elegant.

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

I tried solving it directly using Cauchy-Schwarz inequality.

- 3 years, 7 months ago

Sketch proof: use CS on (1/(a+b)^2 , 1/(a+c)^2 , 1/(b+c)^2 ) with (1,1,1)

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

- 3 years, 7 months ago

I am familiar with the classical inequalities but not with Lagrange Multipliers...I even tried the wiki but couldn't really make it.

- 3 years, 7 months ago

Wait. Are you able to simplify the problem to "Minimize(ab+ac+bc) subject to a,b,c>0 and a+b+c=1" ?

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

- 3 years, 7 months ago

Yes , I could do that.

- 3 years, 7 months ago

[This is wrong, I've made a mistake somewhere, but I can't seem to fix it....]

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).

- 3 years, 7 months ago

Not relevant to this problem,

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

- 3 years, 7 months ago

positive reals.

- 3 years, 7 months ago

Sir, but logically going like weights are the number of times we are using the term , like if $m$ is weight of $a$ , then , it signifies $\underbrace{a + a + a + a + \cdot \cdot \cdot +a}_{\text{x times}}$

- 3 years, 7 months ago

- 3 years, 7 months ago

Sir , I was just reading the wiki on Young's Inequality , and its proof used Weighted AM - GM with weights as positive reals.

- 3 years, 7 months ago

Oh curses. I forgot. Yes, it's positive reals, not "positive integers'. Sorry I got it wrong twice!

- 3 years, 7 months ago

Sir, where am I at fault?

- 3 years, 7 months ago

But as I said in my previous comment , how come the number of ways we are taking a term be a non - natural number?

- 3 years, 7 months ago

Sorry for the delay. I need to read up the articles for AMGM again.

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.

- 3 years, 7 months ago

Sir , in the derivation of AM - GM Inequality , definitely , we'll have to take the function of the particular term like $f(a)$ n number of times to get the power raised in the GM side and that number multiplied in the AM side, so won't that force $n$ to be a positive integer?

- 3 years, 7 months ago

Nope, not necessary. Like I said above, weighted AMGM is a derivation/simpler version of Jensen's, you shouldn't interpret the final weighted form of AMGM as though all the weights must be integers.

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.

- 3 years, 7 months ago

Oh sorry!!I I understood my mistake. Thanks a lot.

- 3 years, 7 months ago

I was reading the wiki on AM - GM Inequality just now and there I saw in the proof that to get the result :

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

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?

- 3 years, 7 months ago

I'll post my complete solution in the weekends...

- 3 years, 7 months ago

Why 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.

- 3 years, 7 months ago