Consider $r \in (0,1)$ and $t_1 <0$ and $t_2 \in \mathbb R.$I want to show that there exists a constant $C_r>0$ such that
$$t_1 (1 - 2 r) + \sqrt{t_1^2 + 4 t_2^2 (1 - r) r} \ge C_r (\vert t_1\vert + \vert t_2\vert).$$
Does anyone know about ways to establish this?
I was able to make some partial progress:One can notice that$$ \sqrt{t_1^2 + 4 t_2^2 (1 - r) r} \ge \frac{-t_1 +2 \vert t_2\vert \sqrt{(1 - r) r}}{\sqrt{2}} .$$
This implies that$$t_1 (1 - 2 r) + \sqrt{t_1^2 + 4 t_2^2 (1 - r) r} \ge -t_1 ( 2 r+2^{-1/2} -1) +\sqrt{2} \vert t_2\vert \sqrt{(1 - r) r} .$$
Thus, the result is true as long as $r>1/2-2^{-3/2}$. However, I am unable to show the result for $r \in (0, 1/2-2^{-3/2}]$.