Note that \[ a^2 = (x-\lambda y)^2 = x^2 +\lambda^2 y^2 - 2\lambda xy \ge x^2 + \lambda^2 y^2 -\lambda x^2 - \lambda y^2. \]Writing the analogous inequalities for the second and the third equation and summing them up, we immediately get the conclusion.

$x=a+\lambda y=a+\lambda(b+\lambda z)=a+\lambda(b+\lambda(c+\lambda x))=a+b\lambda+c\lambda^2+\lambda^3 x\implies x=\frac{a+b\lambda+c\lambda^2}{1-\lambda^3}$ So the system has solution $x=\frac{a+b\lambda+c\lambda^2}{1-\lambda^3}$, $y=\frac{b+c\lambda+a\lambda^2}{1-\lambda^3}$, and $z=\frac{c+a\lambda+b\lambda^2}{1-\lambda^3}$ \begin{align*} x^2+y^2+z^2 &=\frac{(a+b\lambda+c\lambda^2)^2+(b+c\lambda+a\lambda^2)^2+(c+a\lambda+b\lambda^2)^2}{(1-\lambda^3)^2}\\ &=\frac{(a^2+b^2+c^2)(1+\lambda^2+\lambda^4)+2(ab+ac+bc)(\lambda+\lambda^2+\lambda^3)}{(1-\lambda^3)^2}\\ &=\frac{(a^2+b^2+c^2)(1+\lambda+\lambda^2)(1-\lambda+\lambda^2)+2\lambda(ab+ac+bc)(1+\lambda+\lambda^2)}{(1-\lambda^3)^2}\\ &=\frac{(a^2+b^2+c^2)(1-\lambda+\lambda^2)+2\lambda(ab+ac+bc)}{(1-\lambda)^2(1+\lambda+\lambda^2)}\\ &\leq\frac{(a^2+b^2+c^2)(1-\lambda+\lambda^2)+2\lambda(a^2+b^2+c^2)}{(1-\lambda)^2(1+\lambda+\lambda^2)}\quad (\text{Cauchy–Schwarz})\\ &=\frac{(1-\lambda+\lambda^2)+2\lambda}{(1-\lambda)^2(1+\lambda+\lambda^2)}\\ &=\frac{1}{(1-\lambda)^2} \end{align*}