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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!

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  • $\begingroup$ For crying out loud - STOP downvoting comments in the first minute. It's pathetic. $\endgroup$
    – 1729taxi
    Commented Jul 25, 2023 at 9:39

2 Answers 2

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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

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  • $\begingroup$ Ok, does not hold then, thanks ! $\endgroup$
    – Nathan L
    Commented Jul 25, 2023 at 10:53
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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.

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  • $\begingroup$ Got it, makes sense, thanks! $\endgroup$
    – Nathan L
    Commented Jul 25, 2023 at 11:02

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