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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
2 years, 3 months ago

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

- 2 years, 3 months ago

Nice solution. Do you have any more approaches?

- 2 years, 3 months ago

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.

- 2 years, 3 months ago

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

- 2 years, 3 months ago

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

- 2 years, 3 months ago

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

- 2 years, 3 months ago

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

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

- 2 years, 3 months ago

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.

- 2 years, 3 months ago

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

- 2 years, 3 months ago

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

- 2 years, 3 months ago

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

- 2 years, 3 months ago

What if we get maximum when all 4 are equal?

- 2 years, 3 months ago

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