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A classmate consulted me this problem, after a few moment's thought I found it was difficult, so I wish to try my luck here.

Let $z_1,z_2,z_3,z_4\in \mathbb{C}$ such that $|z_1|^2+|z_2|^2=|z_3|^2+|z_4|^2=1$. How to show $$(1-|z_1|^p)^{1/p}\le(1-|z_3|^p)^{1/p}+(1-|z_1\overline{z}_3+z_2\overline{z}_4|^p)^{1/p}, $$ where $p\ge2$ is an integer?

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up vote 2 down vote accepted

Suppose we fix $w_1 = |z_1|$ and $w_3 = |z_3|$. In order to maximize $|z_1\overline{z}_3 + z_2\overline{z}_4|$, we need that the arguments of $z_1\overline{z}_3$ and $z_2\overline{z}_4$ be the same, and in that case the value will be $|z_1||z_3| + |z_2||z_4|$. So we can rewrite the inequality in an equivalent form: $$ (1-w_1^p)^{1/p} \leq (1-w_3^p)^{1/p} + (1-(w_1w_3 + \sqrt{(1-w_1^2)(1-w_3^2)})^p)^{1/p}. $$ Equivalently, $$ |(1-w_1^p)^{1/p} - (1-w_3^p)^{1/p}| \leq (1-(w_1w_3 + \sqrt{(1-w_1^2)(1-w_3^2)})^p)^{1/p}. $$ Writing $w_1 = \sin \alpha$ and $w_3 = \sin \beta$, this is the same as $$ |(1-(\sin \alpha)^p)^{1/p} - (1-(\sin \beta)^p)^{1/p}| \leq (1 - (\cos (\alpha-\beta))^p)^{1/p}. $$ Here $0 \leq \alpha,\beta \leq \pi/2$.

That's as far as I got.

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How did you get the $+$ in the expressions on the left hand side? – Raskolnikov Jun 5 '11 at 19:25
Sorry, corrected. – Yuval Filmus Jun 5 '11 at 19:48
I think what would help is a kind of "continuity relation" for convex/concave functions. $f(x)=(1-x^p)^{1/p}$ for $p\geq 2$ is a concave function. There are often inequalities of the type $|f(x)-f(y)|<f(|x-y|)$ under appropriate restrictions on $x$ and $y$. With a bit of tweaking, I think we could get the end result. – Raskolnikov Jun 5 '11 at 20:01
I was only able to show $|z_2|-|z_4|\le (1-|z_1\overline{z}_3+z_2\overline{z}_4|^p)^{1/p}$. – Sunni Jun 5 '11 at 23:09

From where Yuval left off, it's easy to prove the case $p=2$. In that case

$$|(1-(\sin \alpha)^p)^{1/p} - (1-(\sin \beta)^p)^{1/p}| \leq (1 - (\cos (\alpha-\beta))^p)^{1/p} \; ,$$

reduces to

$$|\cos \alpha - \cos \beta | \leq \sin(\alpha-\beta) \; ,$$

assuming without loss of generality that $\alpha > \beta$. (Note, the case $\alpha=\beta$ is trivial even when $p>2$ anyway). With this assumption, we can also write

$$\cos \beta \leq \cos \alpha + \sin(\alpha-\beta) \; ,$$

or working out the $\sin$ with a formula for subtraction of angles and rearranging

$$\frac{\cos \beta}{1-\sin \beta} \leq \frac{\cos \alpha}{1-\sin \alpha} \; .$$

Thus, proving the case $p=2$ amounts to showing that $\frac{\cos x}{1-\sin x}$ is an increasing function of $x$ over $[0,\pi/2]$, which can be done by looking at its derivative.

I didn't find the case $p>2$ yet, but I still think exploiting the concavity properties of $(1-x^p)^{1/p}$ is a key to solving the problem.

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$$\frac{\cos\,x}{1-\sin\,x}=\cot\left(\frac{\pi}{4}-\frac{x}{2}\right)$$ could have been used as well. – J. M. Nov 22 '11 at 14:02

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