Maximizing the Almost Twin Expressions!

Let us work for $A$ or the first expression.

Let $S=\dfrac{x}{1-yz} + \dfrac{y}{1-zx} + \dfrac{z}{1-xy}$ .

Suppose $xyz \neq 0$ , so that $x,y,z \in (0,1)$ .

Note that $\dfrac{x}{1-yz} = x + \dfrac{xyz}{1-yz}$ . Hence,

$S=x+y+z+xyz \left(\dfrac{1}{1-yz} + \dfrac{1}{1-zx} + \dfrac{1}{1-xy} \right)\quad ....(1)$

Since $1-yz \geq 1 - \dfrac12 \left( y^2 + z^2 \right) = \dfrac12 \left(1+ x^2 \right)$ $= \dfrac12 \left(2x^2 + y^2 + z^2 \right) \geq 2 \sqrt[4]{x^2x^2y^2z^2} = 2x\sqrt{yz}$

we have, using the AM-GM Inequality several times, that

$xyz \left(\dfrac{1}{1-yz} + \dfrac{1}{1-zx} + \dfrac{1}{1-xy} \right)$ $\leq \dfrac{xyz}2 \left( \dfrac1{x\sqrt{yz}} + \dfrac1{y\sqrt{zx}} + \dfrac1{z\sqrt{xy}} \right)$ $\dfrac12 (\sqrt{yz} + \sqrt{zx} + \sqrt{xy}) \leq \dfrac12 \left( \dfrac{x+y}2 + \dfrac{z+x}2 + \dfrac{x+y}2 \right)$ $= \dfrac12(x+y+z) \quad ....(2)$

From (1) and (2), we have, using the Cauchy-Schwarz Inequality, that: $S \leq \dfrac32 (x+y+z) \leq \dfrac32(1^2 + 1^2 + 1^2)^{1/2}(x^2 + y^2 + z^2)^{1/2} = \dfrac{3\sqrt{3}}{2}$

Thus $A = \boxed{\dfrac{3\sqrt{3}}{2}}$ .

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Similarly, by using the concept of Lagrange-Multipliers, we'll obtain $B = \boxed{\sqrt{2}}$ . Please check this solution after a few days so that I'll prove $B=\sqrt{2}$ .

Therefore $A+B \approx 4.0122$ , hence the answer.