Waste less time on Facebook — follow Brilliant.
×

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?

Note by Sambit Senapati
3 years, 9 months ago

No vote yet
4 votes

Comments

Sort by:

Top Newest

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. Paramjit Singh · 3 years, 9 months ago

Log in to reply

@Paramjit Singh Thanks. It was really silly of me to not think of induction. Sambit Senapati · 3 years, 9 months ago

Log in to reply

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 · 3 years, 6 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...