Geometric Inequality

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#1 h Sep 15, 2010, 9:04 PM • 2 Y

Y by Adventure10, Yiyj1

On a semicircle with unit radius four consecutive chords $AB,BC, CD,DE$ with lengths $a, b, c, d$ , respectively, are given. Prove that
$a^2 + b^2 + c^2 + d^2 + abc + bcd < 4.$

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Zhero

#2 h Dec 14, 2010, 11:01 PM • 4 Y

Y by Adventure10, Mango247, Amir Hossein, and 1 other user

Moving $A$ along the semicircle away from $B$ and $D$ along the semicircle away from $C$ until $AD$ is a diameter can only increase $a^2 + b^2 + c^2 + d^2 + abc + bcd$ , so we may suppose that $AD$ is a diameter of the semicircle.

Let $e = CE = \sqrt{c^2 + d^2 - 2cd \cos D}$ . We first claim that $a^2 + b^2 + e^2 + abe = 4$ . $\angle ABE = \angle ACE = \frac{\pi}{2}$ , so $BE = \sqrt{4 - a^2}$ and $AC = \sqrt{4 - e^2}$ . By Ptolemy's theorem, $2b + ae = \sqrt{(4 - a^2)(4 - e^2)}$ , so $4b^2 + 4abe + a^2 e^2 = 16 - 4a^2 - 4e^2 + a^2 e^2$ , which rearranges to $a^2 + b^2 + e^2 + abe = 4$ .

It is therefore sufficient to show that
$a^2 + b^2 + e^2 + abe > a^2 + b^2 + c^2 + d^2 + abc + bcd,$
or
$a^2 + b^2 + c^2 + d^2 - 2cd \cos D + ab \sqrt{c^2 + d^2 - 2cd \cos D} > a^2 + b^2 + c^2 + d^2 + abc + bcd,$
or
$\begin{align} -2cd \cos D + ab \sqrt{c^2 + d^2 - 2cd \cos D} > abc + bcd. \end{align}$
We note that $\angle D \geq \frac{\pi}{2}$ (since the angle is contained in a semicircle), so $\sqrt{c^2 + d^2 - 2cd \cos D} > c$ , whence
$\begin{align*} ab \sqrt{c^2 + d^2 - 2cd \cos D} > abc. \tag{2} \end{align*}$ Furthermore,
$\angle D \&= \pi - \angle CAE = \pi - (\frac{\pi}{2} - \angle CEA) = \frac{\pi}{2} + \angle CEA = \frac{\pi}{2} + (\pi - \angle ABC) = \frac{3\pi}{2} - \angle B,$
so $-\cos D = \sin B$ . By the law of sines, $\frac{AC}{\sin B} = 2$ , so $AC = 2 \sin B = -2 \cos B$ . Because $\angle ABC > \pi$ , $AC > BC = b$ , whence
$\begin{align*} -2cd \cos B > bcd \tag{3}. \end{align*}$
Adding (2) and (3) yields (1), so we are done.

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#3 h Dec 17, 2010, 5:54 AM • 3 Y

Y by Adventure10, Mango247, Amir Hossein

Another solution:

Again, we suppose that $AD$ is a diameter of the semicircle. Let $e = CE$ and $f = CA$ . From a result of the previous post, $a^2 + b^2 + e^2 + abe = 4$ and $f^2 + c^2 + d^2 + fcd = 4$ . Adding these gives $a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + abe + fcd = 8$ . Again, since $\angle B, D > \pi$ , we have $e > c$ and $f > b$ . In addition, $e^2 + f^2 = 4$ by the Pythagorean theorem. Hence, $4 = 8 - (e^2 + f^2) = a^2 + b^2 + c^2 + d^2 + abe + fcd > a^2 + b^2 + c^2 + d^2 + abc + bcd$ .

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