Rearrangement Inequality. Chebyshev’s Inequality

Theorem 1 (Rearrangement inequality)
 
If \( x_1 \), \( x_2 \), \( \dots \), \( x_n \) and \( y_1 \), \( y_2 \), \( \dots \), \( y_n \) are two non-decreasing sequences of real numbers, and if \( \sigma_1 \), \( \sigma_2 \), \( \dots \), \( \sigma_n \) is any permutation of \( \{1,2,\dots, n\} \), then the following inequality holds: \[ x_1y_n+x_2y_{n-1}+\cdots+x_ny_1\leq x_1y_{\sigma_1}+x_2y_{\sigma_2}+\cdots+ x_ny_{\sigma_n}\leq x_1y_1+x_2y_2+\cdots+x_ny_n.\]

Let us show by example how we can prove the inequality between arithmetic and geometric mean using the rearrangement inequality. We will prove it for \( n=4 \), and from there it will be clear how one can generalize the method.

Problem 1
 
If \( a \), \( b \), \( c \), and \( d \) are positive real numbers prove that \[ a^4+b^4+c^4+d^4\geq 4abcd.\]

Theorem 2 (Chebyshev)
 

Let \( a_1\geq a_2\geq\cdots\geq a_n \) and \( b_1\geq b_2\geq\cdots\geq b_n \) be real numbers. Then \begin{eqnarray*} n\sum_{i=1}^n a_ib_i\geq\left(\sum_{i=1}^n a_i \right)\left(\sum_{i=1}^n b_i\right)\geq n\sum_{i=1}^n a_ib_{n+1-i}.\quad\quad\quad\quad\quad (1) \end{eqnarray*} The two inequalities become equalities at the same time when \( a_1=a_2=\cdots=a_n \) or \( b_1=b_2=\cdots=b_n \).

We will prove the following generalization of the above theorem. The left inequality of Theorem 2 follows by substituting \( m_i=\frac1n \) in Theorem 3, and the right inequality follows from the application of the left inequality to the sequences \( (a_i) \) and \( (c_i) \) with \( c_i=-b_{n+1-i} \).

Theorem 3 (Generalized Chebyshev’s Inequality)
 

Let \( a_1\geq a_2\geq\cdots\geq a_n \) and \( b_1\geq b_2\geq\cdots\geq b_n \) be any real numbers, and \( m_1,\dots, m_n \) non-negative real numbers whose sum is \( 1 \). Then \begin{eqnarray*} \sum_{i=1}^n a_ib_im_i\geq\left(\sum_{i=1}^n a_i m_i\right) \left(\sum_{i=1}^n b_im_i\right).\quad\quad\quad\quad\quad (2) \end{eqnarray*} The inequality become an equality if and only if \( a_1=a_2=\cdots=a_n \) or \( b_1=b_2=\cdots=b_n \).

Problem 1
 
If \( a \), \( b \), and \( c \) are positive real numbers, prove the inequality \[ \frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\leq\frac{3(ab+bc+ca)}{2(a+b+c)}.\]

Problem 3
 
Prove that the sum of distances of the orthocenter from the sides of an acute triangle is less than or equal to \( 3r \), where the \( r \) is the inradius.

Problem 4
 
If \( a, b \), and \( c \) are the lengths of the sides of a triangle, \( s \) its semiperimeter, and \( n\geq 1 \) an integer, prove that \[ \frac{a^n}{b+c}+\frac{b^n}{c+a}+ \frac{c^n}{a+b} \geq \left(\frac23\right)^{n-2} \cdot s^{n-1}.\]

Problem 5
 
Let \( 0< x_1 \leq x_2 \leq \cdots \leq x_n \) (\( n\geq 2 \)) and \[ \frac1{1+x_1} + \frac1{1+x_2}+ \cdots + \frac1{1+x_n} = 1.\] Prove that \[ \sqrt{x_1} + \sqrt{x_2} + \cdots + \sqrt{x_n} \geq (n-1) \left( \frac1{\sqrt{x_1}}+ \frac1{\sqrt{x_2}} + \cdots + \frac1{\sqrt{x_n}}\right).\]


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