[latexrefman] \( (x_0+x_1)^2 \leq (x_0)^2+(x_1)^2 \)

[Hefferon, Jim S.]

> I feel uncomfortable with examplifying with an inequation that is not true for all x_0 & x_1 real

I'm not sure why, but you could put a minus on the left and think about it for positive reals, I guess.

Jim

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I believe it important to give many examples, and to underlie the intuition (and sometimes, philosophy) behind definitions and results. This may slow the pace ... for some, in the hope to make it clearer to others.  -- Avi Wigderson

________________________________________
From: Vincent Belaïche <vincent.belaiche at gmail.com>
Sent: Saturday, August 20, 2022 14:06
To: Hefferon, Jim S.
Cc: latexrefman
Subject: Re: \( (x_0+x_1)^2 \leq (x_0)^2+(x_1)^2 \)

⚠ External Sender ⚠


Ooops:  you need x_0 & x_1 to have *OPPOSITE* sign for this to hold

Le sam. 20 août 2022 à 19:00, Vincent Belaïche
<vincent.belaiche at gmail.com> a écrit :
>
> Dear Jim,
>
> I am still propagating your r641, and I noted that you have changed \(
> (x_0+x_1)^2 \) to \( (x_0+x_1)^2 \leq (x_0)^2+(x_1)^2 \) in node
> Subscripts & superscripts.
>
> I feel uncomfortable with examplifying with an inequation that is not true
> for all x_0 & x_1 real (you need x_0 & x_1 to have the same sign for
> this to hold), a strange or controversial example distracts the reader.
>
> I propose either (in order to my descending preference):
>
> - to change it to \( (x_0+x_1)^2 = (x_0)^2+(x_1)^2 + 2 x_0 x_1 \)
> - to revert to \( (x_0+x_1)^2 \)
> - to change it to \( \left(\frac{x_0+x_1}{2}\right)^2} \leq
> \frac{(x_0)^2 + (x_1)^2}{2} \)
> - to change it to \( \sqrt{(x_0+x_1)^2} \leq \sqrt{(x_0)^2}+\sqrt{(x_1)^2} \)
>
>   V.