An inequality involving two complex numbers

Question

Let $z_1, z_2 \in \mathbb C$ and $a,b \in \mathbb{R} \setminus \{0\}$. Prove that

$$|z_1|^2+|z_2|^2-|z_1^2+z_2^2|\le 2\dfrac{|az_1+bz_2|^2}{a^2+b^2}\le |z_1|^2+|z_2|^2+|z_1^2+z_2^2|$$

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Attempt at a solution: Let $z=a+ib$. Then, $2\dfrac{|az_1+bz_2|^2}{a^2+b^2}$ can be simplified into the following

$$\begin{equation} 2\dfrac{|az_1+bz_2|^2}{a^2+b^2} \implies 2\left|\dfrac{\left(\dfrac{z+\bar{z}}{2}\right)z_1+\left(\dfrac{z-\bar{z}}{2i}\right)z_2}{z}\right|^2 \implies\dfrac{\left|\Re(z(z_1+iz_2))\right|^2}{|z|^2} \end{equation}$$

I tried substituting $z=a+ib; z_1=x_1+iy_1; z_2=x_2+iy_2$ but it just became a mess which I think I can't rearrange to make it something useful for proving the inequality. If possible please also provide the geometrical meaning of this inequality.

Answer 1

Let $z_1=x_1+iy_1,z_2=x_2+iy_2$ where $x_1,y_1,x_2,y_2\in\mathbb R$.

Then, $$|z_1|^2+|z_2|^2-|z_1^2+z_2^2|\le 2\dfrac{|az_1+bz_2|^2}{a^2+b^2}\le |z_1|^2+|z_2|^2+|z_1^2+z_2^2|$$ is equivalent to $$x_1^2+y_1^2+x_2^2+y_2^2-\sqrt{(x_1^2-y_1^2+x_2^2-y_2^2)^2+(2x_1y_1+2x_2y_2)^2}\le 2\dfrac{(ax_1+bx_2)^2+(ay_1+by_2)^2}{a^2+b^2}\le x_1^2+y_1^2+x_2^2+y_2^2+\sqrt{(x_1^2-y_1^2+x_2^2-y_2^2)^2+(2x_1y_1+2x_2y_2)^2}$$ So, it is sufficient to prove that $$\sqrt{(x_1^2-y_1^2+x_2^2-y_2^2)^2+(2x_1y_1+2x_2y_2)^2}\ge \left|x_1^2+y_1^2+x_2^2+y_2^2-2\dfrac{(ax_1+bx_2)^2+(ay_1+by_2)^2}{a^2+b^2}\right|$$

Squaring the both sides, $$(x_1^2-y_1^2+x_2^2-y_2^2)^2+(2x_1y_1+2x_2y_2)^2\ge \left(x_1^2+y_1^2+x_2^2+y_2^2-2\dfrac{(ax_1+bx_2)^2+(ay_1+by_2)^2}{a^2+b^2}\right)^2$$ which is equivalent to $$(x_1^2-y_1^2+x_2^2-y_2^2)^2+(2x_1y_1+2x_2y_2)^2-(x_1^2+y_1^2+x_2^2+y_2^2)^2\ge -4(x_1^2+y_1^2+x_2^2+y_2^2)\dfrac{(ax_1+bx_2)^2+(ay_1+by_2)^2}{a^2+b^2}+4\left(\dfrac{(ax_1+bx_2)^2+(ay_1+by_2)^2}{a^2+b^2}\right)^2$$ which is equivalent to $$-(x_1y_2-y_1x_2)^2\ge \dfrac{(ax_1+bx_2)^2+(ay_1+by_2)^2}{a^2+b^2}\cdot\frac{-(ax_2-bx_1)^2-(ay_2-by_1)^2}{a^2+b^2}$$ Multiplying the both sides by $-(a^2+b^2)^2$, $$(a^2+b^2)^2(x_1y_2-y_1x_2)^2\color{red}{\le} ((ax_1+bx_2)^2+(ay_1+by_2)^2)((ay_2-by_1)^2+(bx_1-ax_2)^2)$$ Now, this inequality holds by the Cauchy–Schwarz inequality.