Proof of the Generalised Chebyshev's sum inequality

If $a_1, \ldots ,a_n ; b_1, \ldots ,b_n; \ldots ;k_1, \ldots ,k_n$ are real numbers such that $a_1 \leq \ldots \leq a_n; b_1 \leq \ldots \leq b_n ; \dots ;k_1 \leq \ldots \leq k_n$ then

Prove that

$\frac{\sum_{i=1}^n a_ib_i \ldots k_i}{n}\geq (\frac{\sum_{i=1}^na_i}{n})\cdot (\frac{\sum_{i=1}^nb_i}{n}) \ldots (\frac{\sum_{i=1}^nk_i}{n})$

Recently I read the above inequality in a book stated (without proof) as Generalised Tchebychef's Inequality . Can anyone prove it please?

#Proofs #Math

Easy Math Editor

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Comments

This can be approached by induction on $p$ , the number of sequences involved. Case $p=2$ is the well known Chebyshev's inequality,& can be used as base case. For inductive step, if this is true for $p-1$ , then use:

$\displaystyle \frac{\displaystyle \sum_{i=1}^n a_{1,i} a_{2,i} ... a_{p,i}}{n} \geq \frac{\displaystyle \sum_{i=1}^n a_{p,i}}{n} \frac{\displaystyle \sum_{i=1}^n a_{1,i} a_{2,i}... a_{p-1,i}}{n}$ since $< a_{1,i} \cdot a_{2,i} \cdot... \cdot a_{p-1,i} >$ is also an increasing sequence. Here you may continue by the induction hypothesis. Here, $a_{j,i}$ denotes the $i$ th element of the $j$ th sequence. Use induction as a general method always for generalizing things.

A Brilliant Member - 7 years, 7 months ago

Thanks. It was really silly of me to not think of induction.

Sambit Senapati - 7 years, 7 months ago

Assume that a+b+c <= x+y+z and ab+ac+bc <= xy+xz+yz and abc <= xyz , where a,b,c,x,y,z, are positive numbers . Prove that the previous inequalities hold for a^(1/2) , b^(1/2) , , c^(1/2) , x^(1/2) , y^(1/2) , z^(1/2) !!

For the case n=2 ( I mean only a,b x,y ) , it is clear.

Fozi Dannan - 7 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$