So there is this question of proving where the minimum value occurs of this expression:
$x^2 + \frac{25}{x^2} + 3$ where $x > 0$
After using the A.G.M. inequality:
$\sqrt{xy} \leq \frac{x+y}{2}$
$2\sqrt{xy} \leq x+y$
where $a = x^2$ and $b = \frac{25}{x^2}$, one gets:
$2\sqrt{25} \leq x^2+\frac{25}{x^2}$
$10 \leq x^2+\frac{25}{x^2}$
Now, apparently $\sqrt{25}$ can just simply only take the positive square root. But why don't we consider the negative?
Best Answer
We don't take the negative square root for the same reason we don't take the number $17$ or $\pi^3$ or why we don't take the natural log of the number.
Because that is not what the theorem tells us.
The AM-GM theorem says that $\sqrt{xy} \le \frac{x + y}2$ and that is a theorem about the positive square root.
It is not a theorem about the negative square root. It is not a theorem about the number $17$ or $\pi^3$ or about the natural log.
It is not a theorem about saying "there is a number $u$ so that $u^2 = xy$ and we can conclude $u \le \frac {x + y}2$ but we will have to solve for $u$". It is a theorem that the value $\sqrt{xy}$, which happens to be non -negative, which the theorem gives us on a silver platter and which we are NOT being asked to solve for, that non-negative number $\sqrt{xy}$, is such that $\sqrt{xy} \le \frac {x+y}2$.
Okay, so why don't we have a theorem about the negative square root? Why don't we have a theorem that $\pm\sqrt{xy}\le\frac{x+y}2$? After all that is probably certainly true one would think.
Well, it is true but because $x \ge 0; y \ge 0$ (a necessary condition to the theorem) then we have $-\sqrt{xy} \le 0\le \frac {x+y}2$ and $-\sqrt{xy}\le \frac {x+y}2$ is trivial and not needed. And $\pm \sqrt{xy}\le \frac {x+y} 2$ is a weaker theorem.
If we used that as a theorem, call it $A.M/G.M/B.S$ then we'd get
Okay $x^2 > 0$ and $\frac 25{x^2}> 0$ so by the $A.M/G.M/B.S$ we have $\pm 2\sqrt{x^2*\frac 25{x^2}} = \pm 10 < x^2 + \frac {25}{x^2}$.
But why should we use the $A.M/G.M/B.S$ theorem when we can use the stronger A.M/G.M theorem that says $2\sqrt{25}=10 \le x^2 + \frac {25}{x^2}$.