# Math Help - An inequality

1. ## An inequality

Prove that $|\mathrm{Re}(z)|+|\mathrm{Im}(z)|\leq\sqrt{2}\|z\|$ for all $z\in\mathbb{C}$.

Not very hard, but I wonder to see if there will be nice solutions.

2. Originally Posted by bkarpuz
Prove that $|\mathrm{Re}(z)|+|\mathrm{Im}(z)|\leq\sqrt{2}\|z\|$ for all $z\in\mathbb{C}$.

Not very hard, but I wonder to see if there will be nice solutions.
$a^2 + b^2 \geq 2|a||b|$ is equivalent to $(|a|+|b|)^2 \leq 2(a^2+b^2).$

3. Originally Posted by NonCommAlg
$a^2 + b^2 \geq 2|a||b|$ is equivalent to $(|a|+|b|)^2 \leq 2(a^2+b^2).$
Of course this is the most simple proof.

And still waiting for interesting solutions.

4. If $z=r(\cos\theta+i\sin\theta)\mbox{ with } r=|z|$ then

$|\mathrm{Re}(z)|+|\mathrm{Im}(z)|=r(|\cos\theta|+| \sin\theta|)=r\sqrt{1+|\sin 2\theta|}\leq r\surd2$.

5. Originally Posted by halbard
If $z=r(\cos\theta+i\sin\theta)\mbox{ with } r=|z|$ then

$|\mathrm{Re}(z)|+|\mathrm{Im}(z)|=r(|\cos\theta|+| \sin\theta|)=r\sqrt{1+|\sin 2\theta|}\leq r\surd2$.
My own solution was also similar.
Just set $f(t):=|t|+|\sqrt{1-t^{2}}|$ for $t\in[-1,1]$, and show that $f(t)\leq f(\sqrt{2}/2)=\sqrt{2}$ for all $t\in[0,1]$.
Since $f$ is even, we have $f(\cos(t))\leq\sqrt{2}$ for all $t\in[0,2\pi]$.