Prove: $\left(\sum_{i=1}^na_ix_i^2\right)\left(\sum_{i=1}^n a_iy_i\right)\geq\left(\sum_{i=1}^n a_ix_i\right)\left(\sum_{i=1}^n a_ix_iy_i\right)$

I've been struggling with the following inequality, perhaps because it is not always true after all. Let $$(a_1,\dots,a_n)$$, $$(x_1,\dots,x_n)$$, and $$(y_1,\dots,y_n)$$ be non-negative real numbers. Can we show that the following inequality holds? $$\left(\sum_{i=1}^n a_ix_i^2\right)\left(\sum_{i=1}^n a_iy_i\right)\geq\left(\sum_{i=1}^n a_ix_i\right)\left(\sum_{i=1}^n a_ix_iy_i\right).$$ If it helps, $$x_i\in[0,1]$$ for all $$i$$ and $$x_{i+1}\leq x_i$$. I have tried applying Holder's inequality, and the generalized mean inequality (we can see the $$a_i$$ 's as weights), but those do not deal with two sums on both sides so they do not seem directly applicable. Also, with Holder, some sums should be raised to a power different from one.

Any insight would be much appreciated. Thanks!

• For crying out loud - STOP downvoting comments in the first minute. It's pathetic. Commented Jul 25, 2023 at 9:39

There's a random counter-example :
$$a_1=299.1391904769415$$,
$$a_2=682.8204907409048$$,
$$a_3=164.73223655416868$$,
$$x_1=0.5175401496619542$$,
$$x_2=0.29519300640127977$$,
$$x_3=0.5834232075955864$$,
$$y_1=52.544958893294655$$,
$$y_2=16.028378139651068$$,
$$y_3=73.01122841289978$$,
RHS $$\approx 8317913.971522892$$,
LHS $$\approx 7571490.984361977$$, if I get the things right.
The script

• Ok, does not hold then, thanks ! Commented Jul 25, 2023 at 10:53

Here's a counterexample for $$n = 2$$, with a light explanation:

The LHS can be written as $$a_1^2x_1^2y_1 + a^2x_2^2y_2 + a_1a_2(x_1^2y_2 + x_2^2y_1)$$, while the RHS can be expressed as $$a_1^2x_1^2y_1 + a^2x_2^2y_2 + a_1a_2(x_1x_2(y_1 + y_2))$$, hence LHS $$\geq$$ RHS is equivalent to (when $$a_1a_2 \neq 0$$ anyway) : $$x_1^2y_2 + x_2^2y_1 \overset{??}{\geq} x_1x_2(y_1 + y_2)$$ Now let $$x_1 := 1$$ and $$x_2 := \frac12$$. Then, we have: $$x_1^2 y_2 + x_2^2y_1 \geq x_1x_2(y_1 + y_2) \Longleftrightarrow y_2 + \frac{1}{4} y_1 \geq \frac{1}{2}(y_1 + y_2) \Longleftrightarrow \frac12 y_2 \geq \frac14 y_1$$ But of course that last condition does not always hold, for example you can take $$y_2 = 0$$ and $$y_1 > 0$$, thus the answer to your question is no.

• Got it, makes sense, thanks! Commented Jul 25, 2023 at 11:02