Suppose $a_1,a_2,\ldots,a_n>0 $, where $n>1$ and $\sum_{i=1}^{n}a_i=1$. For each $i=1,2,\ldots,n $, let $b_i=\frac{a^2_i}{\sum\limits_{j=1}^{n}a^2_j}$. Prove that \[\sum_{i=1}^{n}\frac{a_i}{1-a_i}\le \sum_{i=1}^{n}\frac{b_i}{1-b_i} .\] When does equality occur ?
Problem
Source: Paper 1 , Problem 5
Tags: inequalities proposed, inequalities
tarzanjunior
09.01.2017 20:06
\bump Anyone?
BadNick
12.01.2020 19:56
Bump
cefer
13.01.2020 08:56
Let $$S_1 = \sum_{i=1}^{n}a_i \quad \text{and} \quad S_2 = \sum_{i=1}^{n}a_i^2$$Then we can write the given inequality as \[\sum_{i=1}^{n}\frac{a_i}{S_1-a_i}\le \sum_{i=1}^{n}\frac{a_i^2}{S_2-a_i^2} .\]which is equivalent to $$0 \leq \sum_{i=1}^{n}\left(\frac{a_i^2}{S_2-a_i^2} - \frac{a_i}{S_1-a_i}\right) :=X$$Since $$X= \sum_{i=1}^{n}\frac{a_i(a_iS_1 - S_2)}{(S_1-a_i)(S_2-a_i^2)} = \sum_{i=1}^{n}\sum_{j=1}^{n}\frac{a_i(a_ia_j - a_j^2)}{(S_1-a_i)(S_2-a_i^2)} = \sum_{i=1}^{n}\sum_{j=1}^{n}\frac{a_ia_j(a_i - a_j)}{(S_1-a_i)(S_2-a_i^2)}$$we get \begin{align*} 2X &= \sum_{i=1}^{n}\sum_{j=1}^{n}\left(\frac{a_ia_j(a_i - a_j)}{(S_1-a_i)(S_2-a_i^2)} + \frac{a_ja_i(a_j - a_i)}{(S_1-a_j)(S_2-a_j^2)} \right)\\ &=\sum_{i=1}^{n}\sum_{j=1}^{n}a_ia_j(a_i - a_j)\left(\frac{1}{(S_1-a_i)(S_2-a_i^2)} - \frac{1}{(S_1-a_j)(S_2-a_j^2)} \right) \geq 0 \end{align*}The last inequality follows from the fact that function $$f(x) = \frac{1}{(S_1-x)(S_2-x^2)}$$is increasing for all $x < \sqrt{S_2}$.
sqing
13.01.2020 09:12
cefer wrote:
Let
$$S_1 = \sum_{i=1}^{n}a_i \quad \text{and} \quad S_2 = \sum_{i=1}^{n}a_i^2$$Then we can write the given inequality as
\[\sum_{i=1}^{n}\frac{a_i}{S_1-a_i}\le \sum_{i=1}^{n}\frac{a_i^2}{S_2-a_i^2} .\]which is equivalent to
$$0 \leq \sum_{i=1}^{n}\left(\frac{a_i^2}{S_2-a_i^2} - \frac{a_i}{S_1-a_i}\right) :=X$$Since
$$X= \sum_{i=1}^{n}\frac{a_i(a_iS_1 - S_2)}{(S_1-a_i)(S_2-a_i^2)} = \sum_{i=1}^{n}\sum_{j=1}^{n}\frac{a_i(a_ia_j - a_j^2)}{(S_1-a_i)(S_2-a_i^2)} = \sum_{i=1}^{n}\sum_{j=1}^{n}\frac{a_ia_j(a_i - a_j)}{(S_1-a_i)(S_2-a_i^2)}$$we get
\begin{align*}
2X &= \sum_{i=1}^{n}\sum_{j=1}^{n}\left(\frac{a_ia_j(a_i - a_j)}{(S_1-a_i)(S_2-a_i^2)} + \frac{a_ja_i(a_j - a_i)}{(S_1-a_j)(S_2-a_j^2)} \right)\\
&=\sum_{i=1}^{n}\sum_{j=1}^{n}a_ia_j(a_i - a_j)\left(\frac{1}{(S_1-a_i)(S_2-a_i^2)} - \frac{1}{(S_1-a_j)(S_2-a_j^2)} \right) \geq 0
\end{align*}The last inequality follows from the fact that function
$$f(x) = \frac{1}{(S_1-x)(S_2-x^2)}$$is increasing for all $x < \sqrt{S_2}$.