Question

$\textsf{Let x_1,x_2,......,x_{2014} be real positive numbers such that } \displaystyle \sum_{j=1}^{2014} x_j = 1. \\\\\textsf{Determine with proof the smallest constant K such that:}\\\\ K \sum^{2014}_{j=1} \dfrac{x^2_j}{1-x_j} \geq 1.$

Answer 1

Solution :-

As given :-

x₁ ,x₂ ,x₃ ,.....x₂₀₁₄ are real positive number such that

$\sum\limits^{2014}_{j=1} x_j = 1$

Then we have to find out the value of minimum K such that

$K \sum\limits^{2014}_{j=1} \dfrac{x^2_j }{1-x_j} \geq 1$

Now let us suppose that

$\sum\limits^{n}_{j=1} x_j = 1$

Then by solving further :-

$\sum\limits^{n}_{j=1} \dfrac{x^2_j }{1-x_j}$

$= \sum\limits^{n}_{j=1} \dfrac{x^2_j - 1}{1-x_j} + \sum\limits^{n}_{j=1} \dfrac{1}{1-x_j}$

$= \sum\limits^{n}_{j=1} \dfrac{(x_j + 1)(x_j -1)}{-(x_j -1)} + \sum\limits^{n}_{j=1} \dfrac{1}{1-x_j}$

$= \sum\limits^{n}_{j=1} \dfrac{(x_j + 1)\cancel{(x_j -1)}}{-\cancel{(x_j -1)}} + \sum\limits^{n}_{j=1} \dfrac{1}{1-x_j}$

$= \sum\limits^{n}_{j=1} -(x_j + 1) + \sum\limits^{n}_{j=1} \dfrac{1}{1-x_j}$

Now the value of :-

$\sum\limits^{n}_{j=1} -(x_j + 1) = -(n +1)$

And By using AM-HM inequality

$\sum\limits^{n}_{j=1} \dfrac{1}{1-x_j} \\ \\= \sum\limits^{n}_{j=1} \dfrac{\dfrac{1}{1-x_j}}{n} \geq \dfrac{n}{\sum\limits^{n}_{j=1} (1-x_j)}$

$\sum\limits^{n}_{j=1} \dfrac{1}{1-x_j} \geq \dfrac{n^2}{\sum\limits^{n}_{j=1} (1-x_j)}$

So we can say that the minimum value of

$\sum\limits^{n}_{j=1} \dfrac{1}{1-x_j} = \dfrac{n^2}{(n-1)}$

Now the minimum sum of

$\sum\limits^{n}_{j=1} \dfrac{x^2_j }{1-x_j} = \sum\limits^{n}_{j=1} \dfrac{x^2_j - 1}{1-x_j} + \sum\limits^{n}_{j=1} \dfrac{1}{1-x_j}$

$= -(n +1) + \dfrac{n^2}{(n-1)}$

$= \dfrac{-n^2 + 1 + n^2}{(n-1)}$

$= \dfrac{1}{(n-1)}$

So when n = 2014

$\sum\limits^{2014}_{j=1} \dfrac{x^2_j }{1-x_j} \geq \dfrac{1}{(2014 - 1)}$

$\sum\limits^{2014}_{j=1} \dfrac{x^2_j }{1-x_j} \geq \dfrac{1}{2013}$

So the minimum value of K for which

$K \sum\limits^{2014}_{j=1} \dfrac{x^2_j }{1-x_j} \geq 1$

$\Huge{\boxed{\sf{K_{min} = 2013}}}$

Answer 2

Answer:

,x₂ ,x₃ ,.....x₂₀₁₄ are real positive number such that

\sum\limits^{2014}_{j=1} x_j = 1

j=1

∑

2014

x

j

=1

Then we have to find out the value of minimum K such that

K \sum\limits^{2014}_{j=1} \dfrac{x^2_j }{1-x_j} \geq 1K

j=1

∑

2014

1−x

j

x

j

2

≥1

Now let us suppose that

\sum\limits^{n}_{j=1} x_j = 1

j=1

∑

n

x

j

=1

Then by solving further :-

\sum\limits^{n}_{j=1} \dfrac{x^2_j }{1-x_j}

j=1

∑

n

1−x

j

x

j

2

= \sum\limits^{n}_{j=1} \dfrac{x^2_j - 1}{1-x_j} + \sum\limits^{n}_{j=1} \dfrac{1}{1-x_j}=

j=1

∑

n

1−x

j

x

j

2

−1

+

j=1

∑

n

1−x

j

1

= \sum\limits^{n}_{j=1} \dfrac{(x_j + 1)(x_j -1)}{-(x_j -1)} + \sum\limits^{n}_{j=1} \dfrac{1}{1-x_j}=

j=1

∑

n

−(x

j

−1)

(x

j

+1)(x

j

−1)

+

j=1

∑

n

1−x

j

1

= \sum\limits^{n}_{j=1} \dfrac{(x_j + 1)\cancel{(x_j -1)}}{-\cancel{(x_j -1)}} + \sum\limits^{n}_{j=1} \dfrac{1}{1-x_j}=

j=1

∑

n

−

(x

j

−1)

(x

j

+1)

(x

j

−1)

+

j=1

∑

n

1−x

j

1

= \sum\limits^{n}_{j=1} -(x_j + 1) + \sum\limits^{n}_{j=1} \dfrac{1}{1-x_j}=

j=1

∑

n

−(x

j

+1)+

j=1

∑

n

1−x

j

1

Now the value of :-

\sum\limits^{n}_{j=1} -(x_j + 1) = -(n +1)

j=1

∑

n

−(x

j

+1)=−(n+1)

And By using AM-HM inequality

\begin{lgathered}\sum\limits^{n}_{j=1} \dfrac{1}{1-x_j} \\ \\= \sum\limits^{n}_{j=1} \dfrac{\dfrac{1}{1-x_j}}{n} \geq \dfrac{n}{\sum\limit