Some definitions:
A function $f:R^n\rightarrow R$ is convex[strictly convex] if for every $\lambda\in[0,1]$ [$\lambda\in(0,1)$] and for every $x,y$ [$x\neq y$] in $R^n$ we have
$f(\lambda x+(1-\lambda)y)<\lambda f(x)+(1-\lambda)f(y)$ $\qquad$ [$f(\lambda x+(1-\lambda)y)<\lambda f(x)+(1-\lambda)f(y)$]
A convex function $f:R\rightarrow R$ is $ self-concordant$ if $|f^{\prime\prime\prime}(x)|\leq 2f^{\prime\prime}(x)^{\frac{3}{2}}$ for every $x\in domf$.
We say a function $f:R^n\rightarrow R$ is $self-concordant$ if it is $self-concordant$ along every line in its domain, i.e., if the function $\tilde{f}(t) = f(x + tv)$ is a self-concordant function of t for all $x\in domf$ and for all $v$.
Question:
Let $f:R^n\rightarrow R$ be a strictly convex $self-concordant$ function. Suppose $\theta(x)=:(\nabla f(x)^T\nabla^2f(x)^{-1}\nabla f(x))^{\frac{1}{2}}<1$, and define $x^+=x-\nabla^2f(x)^{-1}\nabla f(x)$. Prove that $\theta(x^+)\leq \frac{\theta(x)^2}{(1-\theta(x))^2}$ with all details.(Where the $\nabla^2$ is Hessian and $\nabla$ is gradient)