Show that floating point √x⋅x≥x\sqrt{x \cdot x} \geq x for all long xx.

The idea is simple: Find upper and lower bounds for

$$X := \sqrt{\mathrm{round}(x^2)}$$

and show that $\mathrm{round}(X) = x$.

---

Let $\mathrm{ulp}(x)$ denote the unit of least precision at $x$
and let $E(x)$ and $M(x)$ denote the exponent and mantissa of $x$, i.e.,

$$x = M(x) \cdot 2^{E(x)}$$

with $1 \le M(x) < 2$ and $E(x) \in \mathbb Z$. Define

$$\Delta(x) = \frac{\mathrm{ulp}(x)}x = \frac{\mu \cdot 2^{E(x)}}x = \frac\mu{M(x)}$$

where $\mu=2^{-52}$ is the machine epsilon.

Expressing the rounding function by its relative error leads to

$$X = \sqrt{(1+\epsilon) \cdot x^2} = \sqrt{(1+\epsilon)} \cdot x < \big( 1+\frac\epsilon2 \big) \cdot x$$

We know that $|\epsilon| \le \frac12\Delta(x^2)$ and get (ignoring the trivial case $x=0$)

$$\frac Xx < 1 + \frac{\Delta(x^2)}4 = 1 + \frac\mu{4 M(x^2)}$$

---

By observing $M(x)$ and $M(x^2)$ e.g. over the interval $[1, 4]$,
it can be easily be shown that $\frac{M(x)}{M(x^2)} \le \sqrt2$ which gives us

$$\frac Xx < 1 + \frac{\mu\sqrt2}{4 M(x)}$$

and therefore

$$X < x + \frac{\sqrt2}4 \frac{\mu}{M(x)} \cdot x < x + \frac12 \mathrm{ulp}(x)$$

---

Analogously we get the corresponding lower bound. Just instead of

$$\sqrt{(1+\epsilon)} < \big( 1+\frac\epsilon2 \big)$$

we use something like

$$\sqrt{(1-\epsilon)} > \big( 1 - (1+\epsilon) \cdot \frac\epsilon2 \big)$$

which suffices, since we used a very generous estimate ($\sqrt2/4<\frac12$) in the last step.

---

Because of $|X-x|$ being smaller than $\frac12 \mathrm{ulp}(x)$, $x$ is the double closest to $X$, therefore $\mathrm{round}(X)$ must equal to $x$, q.e.d.