If $|z|<1$, Is $|z+1|+|z-1|<2√2 $ true? If $|z|<1$, Is $|z+1|+|z-1|<2√2 $  true?
My attempt:- 
I got 4 as an upper bound, when I applied triangular inequality on $|z+1|+|z-1|$. I got lower bound as 2,  $2≤|z+1|+|z-1|$.  I randomly pick complex numbers in the given disk. I got the result correct. How to prove or disprove analytically? Please help me.

Taking the square of the magnitude of the LHS:
$$\require{cancel}
\begin{align}
\big| |z+1|+|z-1|\big|^2 &= |z+1|^2+|z-1|^2+2 |z+1||z-1| \\
 &= (z+1)(\bar z+1)+(z-1)(\bar z -1) + 2 \big|(z+1)(z-1)\big| \\
 &= |z|^2+\cancel{z}+\bcancel{\bar z}+1 + |z|^2-\cancel{z}-\bcancel{\bar z} +1  + 2 |z^2-1|\\
 &\le 2 |z|^2 + 2 + 2\big(|z|^2+1\big) \\
 & \lt 8
\end{align}
$$
