I am trying to prove the following inequality
$${\lambda _n}\left( {{X^T}AX} \right) \le {\lambda _n}\left( A \right){\rho ^2}\left( X \right)$$ where $\rho$ is the spectral radius of a matrix and $\lambda_n$ is the smallest eigenvalue. Also, the matrix $A$ is positive definite and $X$ is negative definite.
We may assume that $\lambda_n(X^TAX)>0,$ as otherwise the inequality is obvious. Thus $X$ and $A$ are invertible. For any symmetric matrix $S$ there holds $$\lambda_n(S)=\min\{\langle Su,u\rangle \,:\,\|u\|=1\}$$ Hence $$\lambda_n(X^TAX)=\min\{\langle X^TAXu,u\rangle \,:\,\|u\|=1\}$$ As $X$ is invertible, there exists $v$ such that $$AXv=\lambda_n(A)Xv\quad \|v\|=1$$ Therefore $$\lambda_n(X^TAX)\le \langle X^TAXv,v\rangle =\lambda_n(A) \langle X^TXv,v\rangle \le \lambda_n(A)\|X\|^2$$
By assumption $X$ is negative definite, hence $X$ is symmetric. Thus $\rho(X)=\|X\|.$ Summarizing we get $$\lambda_n(X^TAX)\le \lambda_n(A)\rho(X)^2$$
Remark The assumption $A\ge 0$ is essential. Indeed, let $$X=\begin{pmatrix} -1& 0\\ 0 & 0 \end{pmatrix} ,\quad A=\begin{pmatrix} 0& 0\\ 0 & -1 \end{pmatrix}$$ Then $X^TAX=0$ and $\lambda_2(A)\rho(X)^2=-1.$