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.

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## Comments

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TopNewestBy using \(A.M-H.M\) we get , \[\dfrac{\displaystyle\sum_{cyc}^{a,b,c,d}\left(\frac{a}{b+c+d}+1\right)}{4}\geq\frac{4}{\sum_{cyc}^{a,b,c,d}\left(\dfrac{1}{1+\dfrac{a}{b+c+d}}\right)}\] \[\displaystyle\sum_{cyc}^{a,b,c,d}\left(\dfrac{a}{b+c+d}+1\right)\geq\dfrac{16}{3}\] \[\displaystyle\sum_{cyc}^{a,b,c,d}\left(\dfrac{a}{b+c+d}\right)\geq\dfrac{16}{3}-4\] \[\displaystyle\sum_{cyc}^{a,b,c,d}\left(\dfrac{a}{b+c+d}\right)\geq\dfrac{4}{3}\]

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Nice solution. Do you have any more approaches?

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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.

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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\[\geq \sum_{cyc} a [\dfrac{n^2}{(n-1)\sum_{cyc} a}]-n\] \[=\dfrac{n^2}{n-1} -n\] \[=\dfrac{n}{n-1}\]

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

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

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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

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

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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.

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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

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

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

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What if we get maximum when all 4 are equal?

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