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.

## Comments

Sort by:

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

Log in to reply

– Surya Prakash · 10 months, 3 weeks ago

Nice solution. Do you have any more approaches?Log in to reply

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

Log in to reply

\[\sum_{cyc} \dfrac{a}{\sum_{cyc} a - a}+n -n\] \[=\sum_{cyc} (\dfrac{a}{\sum_{cyc} a - a}+1) -n\] \[=\sum_{cyc} (\dfrac{\sum_{cyc} a}{\sum_{cyc} a - a} -n\] \[=\sum_{cyc} a [\dfrac{1}{\sum_{cyc} a - a}] -n\] 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

Log in to reply

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

Log in to reply

– Nihar Mahajan · 10 months, 2 weeks ago

When I was proving this inequality , I felt a bit restricted , hence gave the generalizationLog in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

– Surya Prakash · 10 months, 3 weeks ago

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

– Dev Sharma · 10 months, 3 weeks ago

'put all 4 equal. To get minimum as 4/3'.. WHY?Log in to reply

– Nihar Mahajan · 10 months, 2 weeks ago

What if we get maximum when all 4 are equal?Log in to reply