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.
 A: the inequality 
$$|z+1|+|z-1|<2√2$$
represent a ellipse with $a=\sqrt{2}$ and $b=\sqrt{2-1}=1$ and center $(0,0)$. So we can write it as:
$$\frac{x^2}{2}+\frac{y^2}{1}<1\to x^2+2y^2<2\quad (1)$$
and
$$|z|<1$$
represent a circle with radius $1$ and center $(0,0)$. So we can write it
$$x^2+y^2<1\quad(2)$$
So you can put the circle inside the ellipse.
Analytically we can write from $(2)$:
$$2x^2+2y^2<2\to \\x^2+2y^2<2-x^2<2$$
which satisfies $(1)$.
A: The unit disk is contained in the elliptic disc $E$ with foci $(\pm1,0)$ and having semiaxes $\sqrt{2}$ and  $1$. Each point $z\in E$ satisfies the given inequality.
A: For each $z$ with $|z|<1$ there is one with $|z|=1$ with a bigger value of $f(z):= |z+1| + |z-1|$. We can write $z=a+bi$ and with $b^2 = 1-a^2$ we get:
$$g(a):=\sqrt{a^2 + 2a + 1 + 1 - a^2} + \sqrt{a^2 - 2a + 1 + 1 - a^2}$$
The optimimum of this in the interval $a\in[0,1]$ has to be $0$, $1$ or having $g'(a)=0$. Checking the values gives us an optimum at $a=0 \rightarrow b=1$ .
$$f(i)= |1+i| +|i-1| = 2\sqrt{2}$$
since $f$ is continuous this bound is tight for $|z|<1$
A: Taking the square of the magnitude of the LHS:
$$\require{cancel}
\begin{align}
\big| |z+1|+|z-1|\big|^2 &= |z+1|^2+|z-1|^2+2 |z+1||z-1| \\
 &= (z+1)(\bar z+1)+(z-1)(\bar z -1) + 2 \big|(z+1)(z-1)\big| \\
 &= |z|^2+\cancel{z}+\bcancel{\bar z}+1 + |z|^2-\cancel{z}-\bcancel{\bar z} +1  + 2 |z^2-1|\\
 &\le 2 |z|^2 + 2 + 2\big(|z|^2+1\big) \\
 & \lt 8
\end{align}
$$
