Minimum value of equations #n+1

Since the two terms of the sum for $A$ are positive, we can use Titu's lemma as follows:

$\begin{aligned} \frac 2{1-x} + \frac 1x & \ge \frac {\left(\sqrt 2+1\right)^2}{1-x+x} \\ \implies A & \ge 3 + 2\sqrt 2 \end{aligned}$

Equality occurs when

$\begin{aligned} \frac 2{1-x} + \frac 1x & = 3 + 2\sqrt 2 \\ (3+2\sqrt 2)x^2 - 2(1+\sqrt 2)x + 1 & = 0 \\ \left((\sqrt 2+1)x-1 \right)^2 & = 0 \\ \implies x & = \frac 1{\sqrt 2 + 1} = \sqrt 2 - 1 \end{aligned}$

Therefore $a+b+c = 3+2+1 = \boxed 6$ .

By rearranging the terms we get $Ax^2-(A-1)x+1=0$ . Since $x$ is real , therefore discriminant of this equation must be non-negative definite. This implies $A^2-6A+1\geq 0$ , or $A\geq {3+2√2}$ . Therefore the minimum value of $A$ is $3+2√2$ , when $x=\dfrac{A-1}{2A}=√2-1$ . Hence $a=3, b=2, c=1$ and $a+b+c=\boxed {6}$