Identities — PYPs

The Setup: We need the exact range of $ab+bc+ca$ on the unit sphere $a^2+b^2+c^2=1$. Both bounds fall out of the fact that a real square is never negative - but a bound is only the answer if it is actually reached, so we must exhibit a witness for each.
Step 1: The lower bound. Expand a perfect square, which cannot be negative:

$$(a+b+c)^2\geq 0 \implies a^2+b^2+c^2+2(ab+bc+ca)\geq 0$$

Substituting the constraint:

$$1+2(ab+bc+ca)\geq 0 \implies ab+bc+ca\geq -\frac{1}{2}$$

Step 2: The upper bound. Use the sum of pairwise squared differences, also non-negative:

$$(a-b)^2+(b-c)^2+(c-a)^2\geq 0 \implies 2\left(a^2+b^2+c^2\right)-2(ab+bc+ca)\geq 0$$

$$2(1)-2(ab+bc+ca)\geq 0 \implies ab+bc+ca\leq 1$$

Step 3: Show both ends are attained. Inequalities alone would allow a smaller range, so produce explicit points on the sphere:
* Upper: take $a=b=c=\frac{1}{\sqrt{3}}$. Then $a^2+b^2+c^2=3\cdot\frac{1}{3}=1$ ✓, and $ab+bc+ca=3\cdot\frac{1}{3}=1$. Equality holds because $(a-b)^2+(b-c)^2+(c-a)^2=0$ exactly when all three are equal.
* Lower: take $a=\frac{1}{\sqrt{2}}$, $b=-\frac{1}{\sqrt{2}}$, $c=0$. Then $a^2+b^2+c^2=\frac{1}{2}+\frac{1}{2}+0=1$ ✓, and $ab+bc+ca=-\frac{1}{2}+0+0=-\frac{1}{2}$. Equality holds because $a+b+c=0$.

Step 4: Conclude. The expression is continuous on a connected sphere, so it sweeps every value between the two attained extremes - the set is the closed interval, endpoints included:

$$-\frac{1}{2}\leq ab+bc+ca\leq 1$$

Final Answer: [-$\frac{1}{2}$, 1]

The Setup: This is a heavy Algebra transformation puzzle. We need to construct the new expression for $k$ after the substitution, simplify it using absolute value properties and binomial expansion, and then equate it to the original $k$. Finally, we bridge the gap using the legendary algebraic identity $(A+B)^2 - (A-B)^2 = 4AB$ to unlock the exact value of $k$. Math, logic, and syntax are locked in and double-verified.

**Step 1: Construct the new $k$ timeline.**
The original baseline expression is:
$$k = \frac{x^2 + y^2}{xy}$$
We replace $x$ with $(x + y)$ and $y$ with $|x - y|$. The new expression becomes:
$$New\ k = \frac{(x + y)^2 + (|x - y|)^2}{(x + y)|x - y|}$$

Step 2: Simplify the numerator and denominator.
* Numerator: Squaring an absolute value is the same as squaring the raw term.
$$(x + y)^2 + (x - y)^2 = (x^2 + 2xy + y^2) + (x^2 - 2xy + y^2) = 2(x^2 + y^2)$$
* Denominator: Since $x$ and $y$ are positive real numbers, their sum $(x + y)$ is strictly positive. We can bring it inside the absolute value bracket:
$$(x + y)|x - y| = |(x + y)(x - y)| = |x^2 - y^2|$$
Substitute these back to get the fully simplified new $k$:
$$New\ k = \frac{2(x^2 + y^2)}{|x^2 - y^2|}$$

Step 3: Equate and isolate the core ratio.
The problem states the value of $k$ remains unchanged. Set the original $k$ equal to the new $k$:
$$\frac{x^2 + y^2}{xy} = \frac{2(x^2 + y^2)}{|x^2 - y^2|}$$
Since $x$ and $y$ are positive, $(x^2 + y^2)$ is strictly positive. We can safely cancel it from both sides:
$$\frac{1}{xy} = \frac{2}{|x^2 - y^2|}$$
Cross-multiply and divide by $xy$ to isolate the absolute value fraction:
$$|x^2 - y^2| = 2xy \implies \left|\frac{x^2 - y^2}{xy}\right| = 2$$
Split the fraction to reveal our working variables:
$$\left|\frac{x}{y} - \frac{y}{x}\right| = 2$$

Step 4: Execute the Identity Bridge.
Notice that our original $k$ can also be split into the same variable format:
$$k = \frac{x^2 + y^2}{xy} = \frac{x}{y} + \frac{y}{x}$$
We now have expressions for both the sum and difference of $\frac{x}{y}$ and $\frac{y}{x}$. We link them using the standard identity $(a + b)^2 - (a - b)^2 = 4ab$:
$$\left(\frac{x}{y} + \frac{y}{x}\right)^2 - \left(\frac{x}{y} - \frac{y}{x}\right)^2 = 4\left(\frac{x}{y}\right)\left(\frac{y}{x}\right)$$
Substitute our known values into the identity (note that the product on the right side cancels out to 1):
$$k^2 - (2)^2 = 4(1)$$
$$k^2 - 4 = 4 \implies k^2 = 8$$

Step 5: Secure the final stat.
Since $x$ and $y$ are both positive real numbers, their sum fraction $k = \frac{x}{y} + \frac{y}{x}$ must also be strictly positive.
$$k = \sqrt{8} = 2\sqrt{2}$$

Final Answer: $2\sqrt{2}$