A well-known version of the Cauchy-Schwarz inequality for positive real numbers (the other one being Titu's lemma) is:
$$ \left( \sum_{i=1}^n u_i v_i \right)^2 \leq \left( \sum_{i=1}^n u_i^2 \right) \left( \sum_{i=1}^n v_i^2 \right).$$
I have seen a few generalisations of C-S including those here but have not yet come across the following one:
$$ \left( \sum_{i=1}^n u_i v_i \right)^3 \leq \left( \sum_{i=1}^n u_i^3 \right) \left( \sum_{i=1}^n v_i^3 \right),\qquad (1)$$
where all the variables are positive real numbers. I can see that we can do:
$$ \left( \sum_{i=1}^n u_i v_i \right)^3 = \left( \sum_{i=1}^n u_i v_i \right)^2 \left( \sum_{i=1}^n u_i v_i \right) \overset{C-S}{\leq} \left( \sum_{i=1}^n u_i^2 \right) \left( \sum_{i=1}^n v_i^2 \right) \left( \sum_{i=1}^n u_i v_i \right),$$
but I do not see how to proceed from here.
So is this cubic version $(\ (1)\ )$ of C-S true, and how do you prove it? And further: is it true if we replace the $3$'s in inequality $\ (1)\ $ with an $\ k \geq 4,\ $ (i.e. the quartic, quintic etc version of $\ (1)\ )\ $ and how are these inequalities proven?
Best Answer
No it's not true. Let $u=(1,1)=v$. Then
$$(\sum_i u_iv_i)^3=2^3$$ but $\sum_i u_i^3=\sum_iv_i^3=2$.