An inequality

**bkarpuz** · March 28th 2009, 12:40 PM

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.

**NonCommAlg** · March 28th 2009, 12:51 PM

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 $(|a|+|b|)^2 \leq 2(a^2+b^2).$

**bkarpuz** · March 28th 2009, 12:59 PM

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.

**halbard** · March 28th 2009, 05:17 PM

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

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

**bkarpuz** · March 29th 2009, 01:14 PM

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 $t\in[0,2\pi]$ .

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