Inequality

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

For a,b,ca, b, c are positive reals, if the infinium (minimum) value of the expression above can be expressed as xy \frac{ x}{y} for positive coprime integers xx and yy, find the value of xy 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
2 years, 8 months ago

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

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

Pi Han Goh - 2 years, 8 months ago

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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,ca, b, c'?

Ankit Kumar Jain - 2 years, 8 months ago

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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)(a,b,c) = (a', b' , c') , then the minimum value also occurs when (a,b,c)=(ka,kb,kc)(a,b,c) = (ka', kb', kc') , where kk is any positive number.

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

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

Pi Han Goh - 2 years, 8 months ago

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

Ankit Kumar Jain - 2 years, 8 months ago

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

Pi Han Goh - 2 years, 8 months ago

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@Pi Han Goh 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.

Ankit Kumar Jain - 2 years, 8 months ago

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@Ankit Kumar Jain 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

Pi Han Goh - 2 years, 8 months ago

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

Ankit Kumar Jain - 2 years, 8 months ago

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@Ankit Kumar Jain I'll post my complete solution in the weekends...

Pi Han Goh - 2 years, 8 months ago

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@Ankit Kumar Jain 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?

Pi Han Goh - 2 years, 8 months ago

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@Pi Han Goh Yes , I could do that.

Ankit Kumar Jain - 2 years, 8 months ago

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

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

Substituting c=1abc = 1 -a-b into this given expression gives a+ba2b2aba + b - a^2 - b^2 - ab .

Let f=a+ba2b2abf = a+b - a^2 - b^2 - ab . Since both a,ba,b are in the open interval (0,1)(0,1) , then ff has no boundary points. So the min/max of ff must occur at its extremal point(s).

Taking the partial derivative of ff with respect to aa, we get fa=12abf_a = 1 - 2a - b .

Similarly, fb=12baf_b = 1 - 2b - a .

At the extremal point(s), fa=fb=0f_a = f_b = 0 . Solving this gives a=b=13f=13a = b= \dfrac13\Rightarrow f = \dfrac13.

Now, we want to prove that f=13f = \dfrac13 is a minimum value. To do so, we apply the second derivative test determinant:

D=faafbb(fab)2, D = f_{aa} f_{bb} - (f_{ab})^2 ,

where faa=2a2f=2f_{aa} = \dfrac{\partial^2}{\partial a^2} f = -2 , fbb=2b2f=2f_{bb} = \dfrac{\partial^2}{\partial b^2} f = -2 and fab=2abf=1f_{ab} = \dfrac{\partial^2}{\partial a \partial b} f = -1 .

This gives D=(2)(2)(1)2=3>0D = (-2)(-2) - (-1)^2 = 3 > 0 with faa(13,13)=13>0f_{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+acab+bc+ac is minimized when a=b=c=13a =b=c= \dfrac13 (under those stated constraints).

Pi Han Goh - 2 years, 8 months ago

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@Pi Han Goh Not relevant to this problem,

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

Ankit Kumar Jain - 2 years, 8 months ago

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@Ankit Kumar Jain positive reals.

Pi Han Goh - 2 years, 8 months ago

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@Pi Han Goh Sir, but logically going like weights are the number of times we are using the term , like if mm is weight of aa , then , it signifies a+a+a+a++ax times\underbrace{a + a + a + a + \cdot \cdot \cdot +a}_{\text{x times}}

Ankit Kumar Jain - 2 years, 8 months ago

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@Ankit Kumar Jain Oh sorry, I misread your question. I thought you're talking about power mean inequality.

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

Pi Han Goh - 2 years, 8 months ago

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

Ankit Kumar Jain - 2 years, 8 months ago

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

Pi Han Goh - 2 years, 8 months ago

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

Ankit Kumar Jain - 2 years, 8 months ago

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@Ankit Kumar Jain 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.

Pi Han Goh - 2 years, 7 months ago

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@Pi Han Goh Sir , in the derivation of AM - GM Inequality , definitely , we'll have to take the function of the particular term like f(a)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 nn to be a positive integer?

Ankit Kumar Jain - 2 years, 7 months ago

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@Ankit Kumar Jain 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.

Pi Han Goh - 2 years, 7 months ago

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

a1+a2+a3++ann(a1a2a3an)1na_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)f(x) as ln(x)ln(x) , then since f"(x)<0f"(x) < 0.

f(a1)+f(a2)+f(a3)++f(an)nf(a1+a2+a3++ann)\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):

f(a1)+f(a1)+f(a1)++f(a1)b_1 times+f(a2)+f(a2)+f(a2)++f(a2)b_2 timesb1+b2f(a1+a1++a1b_1 times+a2+a2++a2b_2 timesb1+b2)\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 bib_i to be a positive integer?

Ankit Kumar Jain - 2 years, 7 months ago

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@Pi Han Goh Oh sorry!!I I understood my mistake. Thanks a lot.

Ankit Kumar Jain - 2 years, 7 months ago

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@Pi Han Goh Sir, where am I at fault?

Ankit Kumar Jain - 2 years, 7 months ago

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Here's the link

Ankit Kumar Jain - 2 years, 8 months ago

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An easy one ans is 5

Md Zuhair - 2 years, 7 months ago

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Please post your complete solution in that case.

Ankit Kumar Jain - 2 years, 7 months ago

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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

Md Zuhair - 2 years, 7 months ago

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@Md Zuhair Okay! :)

Ankit Kumar Jain - 2 years, 7 months ago

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