A question about Young's inequality and complex numbers

[VX10]

TL;DR

Here, I present question about the validity of Young's inequality.

Let $\Omega$ here be $\Omega=\sqrt{-u}$, in which it is not difficult to realize that $\Omega$ is real if $u<0$; imaginary, if $u>0$. Now, suppose further that $u=(a-b)^2$ with $a<0$ and $b>0$ real numbers. Bearing this in mind, I want to demonstrate that $\Omega$ is real. To that end, we must demonstrate that $u<0$, or equivalently,
$$ a^2+b^2-2ab<0$$.
Going through some straightforward algebraic manipulations, we then have
$$ab>\frac{1}{2}\left(a^{2}+b^{2}\right)$$.
Nevertheless, on recalling that $a<0$, we then are led to conclude that
$$ab<\frac{1}{2}\left(a^{2}+b^{2}\right)$$.

Based on the above, I ask:
1. Would that last statement hold true by virtue of Young's inequality?
2. Is there any fallacious step in that given proof?

Thanks in advance.

 

[PeroK]

I have no idea what you are trying to do there, I'm sorry to say.

 

[VX10]

I have no idea what you are trying to do there, I'm sorry to say.

Hi, PeroK. I hope you are doing well. I want to demonstrate that $u<0$.

 

[FactChecker]

Hi, PeroK. I hope you are doing well. I want to demonstrate that $u<0$.

But if $u = (a-b)^2$, isn't $u$ guaranteed to be $\ge 0$?

 

[PeroK]

Hi, PeroK. I hope you are doing well. I want to demonstrate that $u<0$.

And what is $u$?

 

[pasmith]

TL;DR Summary: Here, I present question about the validity of Young's inequality.

Let $\Omega$ here be $\Omega=\sqrt{-u}$, in which it is not difficult to realize that $\Omega$ is real if $u<0$; imaginary, if $u>0$. Now, suppose further that $u=(a-b)^2$ with $a<0$ and $b>0$ real numbers. Bearing this in mind, I want to demonstrate that $\Omega$ is real.

If a and b are real of any sign, then a - b is real and u = (a-b)^2 \geq 0. Hence \Omega is imaginary.

 

[VX10]

But if $u = (a-b)^2$, isn't $u$ guaranteed to be $\ge 0$?

Thanks for commenting FactChecker. This is the point of my doubt. As you can see, when I expand the equation and use the fact that $a<0$, I arrive at the final statement presented above.

 

[VX10]

If a and b are real of any sign, then a - b is real and u = (a-b)^2 \geq 0. Hence \Omega is imaginary.

Hi, pasmith. I hope you are doing well. This is the point of my doubt. As you can see, when I expand the equation and use the fact that $a<0$, I arrive at the final statement presented above. Is the "mathematical development" presented above fallacious? Thanks for commeting.

 

[FactChecker]

Thanks for commenting FactChecker. This is the point of my doubt. As you can see, when I expand the equation and use the fact that $a<0$, I arrive at the final statement presented above.

Any real number, when squared, is positive. It is pointless to look at how that real number was obtained.

 

[Office_Shredder]

Hi, pasmith. I hope you are doing well. This is the point of my doubt. As you can see, when I expand the equation and use the fact that $a<0$, I arrive at the final statement presented above. Is the "mathematical development" presented above fallacious? Thanks for commeting.

There's no development. To show $u<0$ you decided it was equivalent to $ab> 0.5(a^2+b^2)$, but you concluded that actually the opposite is true. This means you proved $u>0$.
But this is all nonsense, since that last step relies on $a^2$ and $b^2$ being positive, which is something you didn't want to assume to begin with.

 

[PeroK]

Hi, pasmith. I hope you are doing well. This is the point of my doubt. As you can see, when I expand the equation and use the fact that $a<0$, I arrive at the final statement presented above. Is the "mathematical development" presented above fallacious? Thanks for commeting.

As I said before your steps make little or no sense. $(a-b)^2 \ge 0$ for all real $a, b$. It's not clear how or why you think you have shown that $(a - b)^2 < 0$.

 

[Mark44]

Any real number, when squared, is positive.

Or zero...

 

[VX10]

There's no development. To show $u<0$ you decided it was equivalent to $ab> 0.5(a^2+b^2)$, but you concluded that actually the opposite is true. This means you proved $u>0$.
But this is all nonsense, since that last step relies on $a^2$ and $b^2$ being positive, which is something you didn't want to assume to begin with.

Thanks for commeting. Now, it became clear to me. Thanks again.

 

[Mark44]

Let $\Omega$ here be $\Omega=\sqrt{-u}$, in which it is not difficult to realize that $\Omega$ is real if $u<0$; imaginary, if $u>0$. Now, suppose further that $u=(a-b)^2$ with $a<0$ and $b>0$ real numbers. Bearing this in mind, I want to demonstrate that $\Omega$ is real.

To summarize what others have said, if $u = (a - b)^2$, with a and be being any real numbers, then $u \ge 0$. So $\Omega$ is zero if a = b or is imaginary otherwise. Period.