Solution for approximating square roots? I just saw a post on instagram which said that
$$\sqrt{x}\approx \frac{x+y}{2\sqrt{y}}$$
I tried it out on a few values, and surprisingly, it came within 1 decimal point of the actual answer. Is there a reason for this or is it coincidental?
y is the closest perfect square to x
 A: This is because, for $x\approx y$:
$$\sqrt{xy}\approx \frac{x+y}{2}$$
This is a consequence of the AM-GM Inequality, which states that:
$$\frac{x+y}{2}\geq\sqrt{xy}$$
For all $x, y\geq 0$, with equality occurring if and only if $x=y$.
A: Using Taylor series around $y=x$ (assuming them to be positive), we have
$$\sqrt{x\,y}=x+\frac{y-x}{2}-\frac{(y-x)^2}{8 x}+\frac{(y-x)^3}{16 x^2}+O\left((y-x)^4\right)$$ and,sice it is an alternating series, you can build a lot of ineaqualities.
A: Google AM–GM inequality.
If we square any real number $z$, which can be expressed as a difference of two other real numbers $x$ and $y$, the result is always greater or equal $0$, this means $\color{blue}{z^2 \geq 0}$, assume $z = x-y$, it follows $$\color{blue}{z^2} = (x-y)^2 = x^2 \color{red}{-2xy} + y^2 = x^2 \color{red}{+ 2xy} + y^2 \color{red}{- 4xy} =(x+y)^2-4xy\color{blue}{\geq 0} $$
$$\Rightarrow (x+y)^2-4xy\geq 0 \Leftrightarrow(x+y)^2\geq 4xy \Rightarrow x+y \geq \sqrt{4xy} \Leftrightarrow \frac{x+y}{2\sqrt{y}} \geq \sqrt{x}$$
The result is:
$$\sqrt{x} \leq \frac{x+y}{2\sqrt{y}}  $$
If $x = y$, then
$$\sqrt{x} =\frac{x+y}{2\sqrt{y}}  $$
Example: $$\sqrt{4} =\frac{4+4}{2\sqrt{4}} = 2$$
A: Your approximation is for lower disadvantages, there's a better approximation for the Square root function, check here for it's derivation
$$\sqrt{x} \approx \frac{ x+k^2+k}{2k+1}$$
Say $k^2+k = y$, $k^2+k-y = 0$
$$k = \frac{ -1 \pm \sqrt{1+4y} }{2}$$
$$\sqrt{x} = \approx \frac{x+y}{2k+1}$$
$$2k+1 = \sqrt{1+4y}$$
$$2k+1 = 2\sqrt{y+\frac{1}{4}}$$
$$2k+1 \approx 2\sqrt{y}$$
$$\sqrt{x} = \approx \frac{ x+y}{2\sqrt{y}}$$
