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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

\[\dfrac{a}{b+c+d} + \dfrac{b}{a+c+d} + \dfrac{c}{a+b+d} + \dfrac{d}{a+b+c}\]

Details and Assumptions:

  • \(a\), \(b\), \(c\) and \(d\) are positive real numbers.

Try to make different approaches.


Try more proof problems at Olympiad Proof Problems.

Note by Surya Prakash
10 months, 3 weeks ago

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By using \(A.M-H.M\) we get , \[\dfrac{\displaystyle\sum_{cyc}^{a,b,c,d}\left(\frac{a}{b+c+d}+1\right)}{4}\geq\frac{4}{\sum_{cyc}^{a,b,c,d}\left(\dfrac{1}{1+\dfrac{a}{b+c+d}}\right)}\] \[\displaystyle\sum_{cyc}^{a,b,c,d}\left(\dfrac{a}{b+c+d}+1\right)\geq\dfrac{16}{3}\] \[\displaystyle\sum_{cyc}^{a,b,c,d}\left(\dfrac{a}{b+c+d}\right)\geq\dfrac{16}{3}-4\] \[\displaystyle\sum_{cyc}^{a,b,c,d}\left(\dfrac{a}{b+c+d}\right)\geq\dfrac{4}{3}\] Shivam Jadhav · 10 months, 3 weeks ago

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@Shivam Jadhav Nice solution. Do you have any more approaches? Surya Prakash · 10 months, 3 weeks ago

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Generalization:

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

When \(n=3\) we get Nesbitt's inequality and when \(n=4\) we get the inequality of this note. Nihar Mahajan · 10 months, 2 weeks ago

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@Nihar Mahajan Proof:

\[\sum_{cyc} \dfrac{a}{\sum_{cyc} a - a}+n -n\] \[=\sum_{cyc} (\dfrac{a}{\sum_{cyc} a - a}+1) -n\] \[=\sum_{cyc} (\dfrac{\sum_{cyc} a}{\sum_{cyc} a - a} -n\] \[=\sum_{cyc} a [\dfrac{1}{\sum_{cyc} a - a}] -n\] Titu's Lemma/Cauchy-Shwarz in engel form\[\geq \sum_{cyc} a [\dfrac{n^2}{(n-1)\sum_{cyc} a}]-n\] \[=\dfrac{n^2}{n-1} -n\] \[=\dfrac{n}{n-1}\] Sualeh Asif · 10 months, 2 weeks ago

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@Nihar Mahajan @Nihar Mahajan nice idea..... how did you reach at this generalization? Dev Sharma · 10 months, 2 weeks ago

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@Dev Sharma When I was proving this inequality , I felt a bit restricted , hence gave the generalization Nihar Mahajan · 10 months, 2 weeks ago

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So, here is my solution.

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

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

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

It implies that

\[f(a) + f(b) + f(c) + f(d) \geq 4f\left(\dfrac{a+b+c+d}{4}\right) = 4 f \left(\dfrac{1}{4}\right) = \dfrac{4}{3}\]. Surya Prakash · 10 months, 2 weeks 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 · 10 months, 2 weeks 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 · 10 months, 3 weeks ago

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@Aakash Khandelwal How can you say that minimum occurs when all the four quantities are equal? You should prove that. Surya Prakash · 10 months, 3 weeks ago

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@Aakash Khandelwal 'put all 4 equal. To get minimum as 4/3'.. WHY? Dev Sharma · 10 months, 3 weeks ago

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@Aakash Khandelwal What if we get maximum when all 4 are equal? Nihar Mahajan · 10 months, 2 weeks ago

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