The expression $1 + x^2 +(-T_px+y)^2 +z^2$ is bounded below by a constant multiple of $(1+x^2+y^2+z^2)$ Suppose $T_p > 0$. Is there a simply way to show that 
$1 + x^2 +(-T_px+y)^2 +z^2 \geq C (1+x^2+y^2+z^2)$, for all $(x,y,z) \in \mathbb R^3$, where $C>0$.  

We can show that for a certain positive constant $C$ 
$$ x^2+(-T_p x+y)^2+z^2+w^2 \geq C(x^2+y^2+z^2+w^2) \tag{1} $$
holds for any $(x,y,z,w)\in\mathbb{R}^4$. We just need that the eigenvalues of the symmetric matrix
$$ M=\left(\begin{array}{cccc}T_p^2+1-C & -T_p & 0 & 0 \\ -T_p & 1-C & 0 & 0 \\ 0 & 0 & 1-C & 0 \\ 0 & 0 & 0 & 1-C\end{array}  \right) $$
are non-negative. This is equivalent to asking that:
$$ C\in(0,1],\qquad (T_p^2+1-C)(1-C)-T_p^2 \geq 0 \tag{2} $$
hence
$$ C = \frac{2+T_p^2-T_p\sqrt{4+T_p^2}}{2}\in (0,1]\tag{3}$$
is an effective choice.
