Olympiad proof Problem - Day 1

From this day onwards I am going to post some proof problems.So that you can use these as practice problems for Olympiads. Let us start from basic level (not very basic though).


Find the minimum value of the expression below

ab+c+d+ba+c+d+ca+b+d+da+b+c\dfrac{a}{b+c+d} + \dfrac{b}{a+c+d} + \dfrac{c}{a+b+d} + \dfrac{d}{a+b+c}

Details and Assumptions:

  • aa, bb, cc and dd are positive real numbers.

Try to make different approaches.


Try more proof problems at Olympiad Proof Problems.

Note by Surya Prakash
3 years, 11 months ago

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By using A.MH.MA.M-H.M we get , cyca,b,c,d(ab+c+d+1)44cyca,b,c,d(11+ab+c+d)\dfrac{\displaystyle\sum_{cyc}^{a,b,c,d}\left(\frac{a}{b+c+d}+1\right)}{4}\geq\frac{4}{\sum_{cyc}^{a,b,c,d}\left(\dfrac{1}{1+\dfrac{a}{b+c+d}}\right)} cyca,b,c,d(ab+c+d+1)163\displaystyle\sum_{cyc}^{a,b,c,d}\left(\dfrac{a}{b+c+d}+1\right)\geq\dfrac{16}{3} cyca,b,c,d(ab+c+d)1634\displaystyle\sum_{cyc}^{a,b,c,d}\left(\dfrac{a}{b+c+d}\right)\geq\dfrac{16}{3}-4 cyca,b,c,d(ab+c+d)43\displaystyle\sum_{cyc}^{a,b,c,d}\left(\dfrac{a}{b+c+d}\right)\geq\dfrac{4}{3}

Shivam Jadhav - 3 years, 11 months ago

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Nice solution. Do you have any more approaches?

Surya Prakash - 3 years, 11 months ago

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Generalization:

i=1naij=1najainn1\Large{\sum_{i=1}^n}\large{ \dfrac{a_i}{\small{\displaystyle\sum_{j=1}^{n}a_{j}-a_i}}\geq \dfrac{n}{n-1}}

When n=3n=3 we get Nesbitt's inequality and when n=4n=4 we get the inequality of this note.

Nihar Mahajan - 3 years, 11 months ago

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Proof:

cycacycaa+nn\sum_{cyc} \dfrac{a}{\sum_{cyc} a - a}+n -n =cyc(acycaa+1)n=\sum_{cyc} (\dfrac{a}{\sum_{cyc} a - a}+1) -n =cyc(cycacycaan=\sum_{cyc} (\dfrac{\sum_{cyc} a}{\sum_{cyc} a - a} -n =cyca[1cycaa]n=\sum_{cyc} a [\dfrac{1}{\sum_{cyc} a - a}] -n Titu's Lemma/Cauchy-Shwarz in engel formcyca[n2(n1)cyca]n\geq \sum_{cyc} a [\dfrac{n^2}{(n-1)\sum_{cyc} a}]-n =n2n1n=\dfrac{n^2}{n-1} -n =nn1=\dfrac{n}{n-1}

Sualeh Asif - 3 years, 11 months ago

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@Nihar Mahajan nice idea..... how did you reach at this generalization?

Dev Sharma - 3 years, 11 months ago

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When I was proving this inequality , I felt a bit restricted , hence gave the generalization

Nihar Mahajan - 3 years, 11 months ago

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So, here is my solution.

The given expression is homogeneous equation of degree 00. So, we can make a constraint a+b+c+d=1a+b+c+d=1. And this constraint gives us that 0<a,b,c,d<10 < a,b,c,d <1.

The expression transforms into a,b,c,da1a\sum_{a,b,c,d} \dfrac{a}{1-a}

Let f(x)=x1xf(x) = \dfrac{x}{1-x}. Then we get that f(x)=2(x1)3>0f''(x) = -2(x-1)^{-3} > 0 for all x<1x<1.

It implies that

f(a)+f(b)+f(c)+f(d)4f(a+b+c+d4)=4f(14)=43f(a) + f(b) + f(c) + f(d) \geq 4f\left(\dfrac{a+b+c+d}{4}\right) = 4 f \left(\dfrac{1}{4}\right) = \dfrac{4}{3}.

Surya Prakash - 3 years, 11 months ago

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We can also solve it using Titu's Lemma. I am currently out of station. I would post the proof as soon as I return home.

Nihar Mahajan - 3 years, 11 months ago

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1) put all 4 equal. To get minimum as 4/3. 2)put deniminator of all fractions as w,x,yand z. Now find each a,b,c and in terms of w,x,y and z. All these (w,x,y and z )are positive real no.s. now apply AM>=GM . 3) Rather finding out each a to d in terms of w,x,y and z , one can add +1 to each term take (a+b+c+d) common apply AM>=GM . Later substract 4 from answer

Aakash Khandelwal - 3 years, 11 months ago

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'put all 4 equal. To get minimum as 4/3'.. WHY?

Dev Sharma - 3 years, 11 months ago

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How can you say that minimum occurs when all the four quantities are equal? You should prove that.

Surya Prakash - 3 years, 11 months ago

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What if we get maximum when all 4 are equal?

Nihar Mahajan - 3 years, 11 months ago

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