Verify that $\sqrt{2}\left\| z \right\| \ge \left|\Re(z)\right| + \left|\Im(z)\right|$

Question

Verify that $\sqrt{2}\left\| z \right\| \ge \left|\Re(z)\right| + \left|\Im(z)\right|.$

I started off noting that $z=x+iy$ and that $Re(z)=x$ and $Im(z)=y$

Then I know that I have to square both sides, giving me

$$2(x^2 +y^2) \ge x^2+2|xy|+y^2$$

I'm not sure what to do after this.

Answer 1

Let $z=x+iy$. We have $(x-y)^2\geq 0$, therefore \begin{align} 2|z|^2=2\big(|x|^2+|y|^2\big)&\geq |x|^2+|y|^2+2|x||y|\\ &=\big(|x|+|y|\big)^2\\ &=\big(|\Re(z)|+|\Im(z)|\big)^2 \end{align} so we have that $$\sqrt2\,|z|\:\geq|\:\Re (z)|+|\Im (z)|\,.$$

Answer 2

Hint:

Let $z = a + ib$ Then we have:

$$\sqrt{2}\sqrt{a^2+b^2} \geq |a| + |b| $$ $$\sqrt{2(a^2+b^2)} \geq |a| + |b| $$ $$ 2(a^2+b^2) \geq (|a| + |b|)^2$$

Try to simplify it so LHS is an expression, and the right hand side is $0$. Notice what the left hand side is and confirm the equality.