# An Inequality with Areas, Norms, and Complex Numbers

### Solution 1

Recollect that

(*)

$\sin 3\alpha=3\sin\alpha-4\sin^3\alpha\,$

and consider two points $A=(a,b)\,$ and $B=(c,d)\,$ in the plane with the origin $O=(0,0).\,$ The area $[\Delta ABO]\,$ can be expressed in two ways, viz.,

$2[\Delta ABO]=|ad-bc|,\,\text{and}\\ 2[\Delta ABO]=\sqrt{a^2+b^2}\sqrt{c^2+d^2}\sin\alpha,$

where $\alpha\,$ is the angle at vertex $O\,$ of the triangle. It follows that

$\displaystyle \sin\alpha = \frac{|ad-bc|}{\sqrt{(a^2+b^2)(c^2+d^2)}}.$

By noticing that $|\sin 3\alpha |\le 1\,$ and substituting into (*), we obtain

\displaystyle \begin{align} 1&\ge 3\frac{|ad-bc|}{\sqrt{(a^2+b^2)(c^2+d^2)}}-4\left(\frac{|ad-bc|}{\sqrt{(a^2+b^2)(c^2+d^2)}}\right)^3\\ &=\frac{|ad-bc|(3(a^2+b^2)(c^2+d^2)-4(ad-bc)^2)}{\left((a^2+b^2)(c^2+d^2)\right)^{\frac{3}{2}}}, \end{align}

which is the required inequality.

### Solution 2

Let $Z_1=a+ib,\,$ $z_2=d+ic.\,$ Then $z_1z_2=(ad-bc)+i(ac+bd),\,$ $ad-bc=\frac{1}{2}(z_1z_2-\overline{z_1z_2}),\,$ $ac+bd=\frac{1}{2}(z_1z_2+\overline{z_1z_2}).\,$ Also, $|z_1|^2=a^2+b^2\,$ and $|z_2|^2=c^2+d^2.\,$ Now,

\displaystyle \begin{align} 3(a^2+b^2)(c^2+d^2)&-4(ad-bc)^2=3z_1\overline{z_1}z_2\overline{z_2}-\frac{4}{4}(z_1z_2+\overline{z_1z_2})^2\\ &=-[z_1^2z_2^2+\overline{z_1}^2\overline{z_2}^2+2z_1z_2\overline{z_1z_2}-3z_1z_2\overline{z_1z_2}]\\ &=-[z_1^2z_2^2+\overline{z_1}^2\overline{z_2}^2-z_1z_2\overline{z_1z_2}]\\ \end{align}

Set

\displaystyle \begin{align} Num &= -\frac{1}{2}(z_1z_2+\overline{z_1z_2})[z_1^2z_2^2+\overline{z_1}^2\overline{z_2}^2-z_1z_2\overline{z_1z_2}]\\ &=-\frac{1}{2}\left[(z_1z_2)^3-(\overline{z_1z_2})^3\right] \end{align}

so that

\displaystyle \begin{align} |Num|&\le\frac{1}{2}\left|(z_1z_2)^3-(\overline{z_1z_2})^3\right|\\ &\le\frac{1}{2}\left[|z_1z_2|^3+|\overline{z_1z_2}|^3\right]\\ &=|z_1z_2|^3=\left((a^2+b^2)(c^2+d^2)\right)^{\frac{3}{2}}=Den, \end{align}

implying $\displaystyle \frac{Num}{Den}\le \frac{|Num|}{Den}\le 1.$

### Solution 3

Set $z_1=a+ib=r_1(\cos\theta+i\sin\theta)\,$ and $z_2=d+ic=r_2(\cos\phi+i\sin\phi).\,$

\begin{align} z_1z_2 &=(ad-bc)+i(ac+bd)\\ &=r_1r_2[\cos (\theta +\phi )+i\sin (\theta +\phi )]. \end{align}

Also, $|z_1|=r_1\,$ and $|z_2|=r_2.\,$ Now,

\displaystyle \begin{align} &\frac{(ad-bc)(3(a^2+b^2)(c^2+d^2)-4(ad-bc)^2)}{\left((a^2+b^2)(c^2+d^2)\right)^{\frac{3}{2}}}\\ &=\frac{r_1r_2\cos (\theta+\phi)[3r_1^2r_2^2-4r_1^2r_2^2\cos^2(\theta+\phi)]}{r_1^3r_2^3}\\ &=3\cos (\theta+\phi)-4\cos^3(\theta+\phi)\\ &=-\cos 3(\theta+\phi)\le 1. \end{align}

### Solution 4

Let $(a^2+b^2)(c^2+d^2)=x^2,\,$ $x\gt 0.\,$ We need to prove that

\displaystyle \begin{align} &\frac{ad-bc)(3x^2-4(ad-bc)^2}{x^3}\le 1\,\Longleftrightarrow\\ &x^3-3x^2(ad-bc)+4(ad-bc)^3\ge 0\,\Longleftrightarrow\\ &x^3+(ad-bc)^3-3(ad-bc)[x^2-(ad-bc)^2]\ge 0\,\Longleftrightarrow\\ &(x+ad-bc)(x^2-x(ad-bc)+(ad-bc)^2)\\ &(x+ad-bc)[x^2-4x(ad-bc)+4(ad-bc)^2]\,\Longleftrightarrow\\ &\qquad\qquad\qquad -3(ad-bc)[x^2-(ad-bc)^2]\ge 0\,\Longleftrightarrow\\ &(x+ad-bc)(x-2(ad-bc))^2\ge 0, \end{align}

which is true, for $x\ge bc-ad\,$ since $(ac+bd)^2\ge 0,\,$ for $a,b,c,d\in\mathbb{R}.$

### Solution 5

From $(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2),$

\begin{align} 3(a^2+b^2)(c^2+d^2)&-4(ad-bc)^2\\ &=3[(ac+bd)^2+(ad-bc)^2]-4(ad-bc)^2]\\ &\overset{(1)}{=}3(ac+bd)^2-(ad-bc)^2\\ &=3|ac+bd|^2-|ad-bc|^2. \end{align}

Case 1: $\mathbf{ac+bd\ne 0,\,ad-bc\ne 0}$

From the diagram, $|ac+bd|=p\cos\theta,\,(2)\,$ and $|ad-bc|=p\sin\theta,\,(3),\,$ where $p=\sqrt{(a^2+b^2)(c^2+d^2)}.\,$ It follows that

\displaystyle \begin{align} LHS &= \frac{(ad-bc)(3p^2\cos^2\theta-p^2\sin^2\theta)}{p^3},\,\text{using (1), (2), (3)}\\ &\overset{(4)}{=}\frac{(ad-bc)(3\cos^2\theta-\sin^2\theta)}{p}. \end{align}

Now, according as $ad-bc\ge 0\,$ or $ad-bc\le 0,\,$ $ad-bc=\pm|ad-bc|,\,(5).\,$ In any event, $\displaystyle \frac{|d-bc|}{p}=\sin\theta.\,$ Such that, using (4) and (5),

\begin{align} LHS &=\pm\sin\theta(3\cos^2\theta-\sin^2\theta)\\ &=\pm\sin\theta(3(1-\sin^2\theta)-\sin^2\theta)\\ &=\pm(3\sin\theta-4\sin^3\theta)\\=\pm\sin3\theta. \end{align}

Since $|\sin 3\theta |\le 1,\,$ $|LHS|\le 1.$

Case 2: $\mathbf{ad-bc=0}$

Then $ac+bd\ne 0\,$ and $LHS=0\le 1.$

Case 3: $\mathbf{ac+bd=0}$

Then $ad-bc=0\,$ and $\displaystyle \displaystyle LHS=\frac{-(ad-bc)}{|ad-bc|}=\pm 1\le 1.$

### Acknowledgment

This is Problem SP060 from the 2017 Spring issue of the Romanian Mathematical Magazine, proposed by Dan Sitaru. Solutions 2 and 3 are by Ravi Prakash, Solution 4 is by Soumitra Mandal; Solution 5 is by Soumava Chakraborty (all India).