2

Let $ABCD$ be a cyclic quadrilateral with sides $a$, $b$, $c$ and $d$. Denote $s$ the semiperimeter and let $\angle{DAB}=\alpha$, $\angle{ABC}=\beta$, $\angle{BCD}=\gamma$ and $\angle{CDA}=\delta$. Then the following inequality holds

$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\gamma}{2}}\tan{\frac{\delta}{2}}+\tan{\frac{\delta}{2}}\tan{\frac{\alpha}{2}}\geq4.\tag{1}$$

Proof. Substituting from the half-angle formula for the tangent (see $(2)$ here) we have that $$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}=\sqrt{\frac{(s-a)(s-d)}{(s-b)(s-c)}}\cdot{\sqrt{\frac{(s-a)(s-b)}{(s-c)(s-d)}}}=\frac{s-a}{s-c}.$$ Similarly, $$\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}=\frac{s-b}{s-d}\qquad\tan{\frac{\gamma}{2}}\tan{\frac{\delta}{2}}=\frac{s-c}{s-a}\qquad\tan{\frac{\delta}{2}}\tan{\frac{\alpha}{2}}=\frac{s-d}{s-b}$$ Thus, the left-hand side of $(1)$ can be rewritten as follows $$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\gamma}{2}}\tan{\frac{\delta}{2}}+\tan{\frac{\delta}{2}}\tan{\frac{\alpha}{2}}=\frac{s-a}{s-c}+\frac{s-b}{s-d}+\frac{s-c}{s-a}+\frac{s-d}{s-b}.$$ But, $$\frac{s-a}{s-c}+\frac{s-b}{s-d}+\frac{s-c}{s-a}+\frac{s-d}{s-b}=\frac{a-c}{s-a}+\frac{b-d}{s-b}+\frac{c-a}{s-c}+\frac{d-b}{s-d}+4.$$ Since $\frac{a-c}{s-a}+\frac{c-a}{s-c}=\frac{4(a-c)^2}{(-a+b+c+d)(a+b-c+d)}$, and similarly for $\frac{b-d}{s-b}+\frac{d-b}{s-d}$, then $\frac{a-c}{s-a}+\frac{b-d}{s-b}+\frac{c-a}{s-c}+\frac{d-b}{s-d}$ is positive. Hence,

$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\gamma}{2}}\tan{\frac{\delta}{2}}+\tan{\frac{\delta}{2}}\tan{\frac{\alpha}{2}}\geq4.$$

Notice equality holds when $ABCD$ is rectangular.

A huge list of inequalities can be seen at Cut-the-knot.org.

Questions: a) Is there a simpler way to prove $(1)$? b) Is $(1)$ known?

Blue
  • 75,673
Emmanuel José García
  • 807
  • 3
    Use the Arithmetic-Geometric Mean Inequality and the fact that $\alpha/2$ and $\gamma/2$ are complements (hence have reciprocal tangents), as are $\beta/2$ and $\delta/2$. – Blue Jun 20 '22 at 02:42
  • @Blue: Isn't the proof of the AM-GM inequality more complicated than my proof of $(1)$? I don't know, but I feel like cutting bread with a saw using AM-GM inequality. – Emmanuel José García Jun 20 '22 at 12:00
  • Well, the two-term AM-GM is dead-simple. So, if we write $x$, $y$, $1/x$, $1/y$ for the tangents of the successive half-angles, the target sum is $$xy+\frac{y}{x}+\frac1{xy}+\frac{x}{y}=\left(x+\frac1x\right)\left(y+\frac1y\right)\geq 2\cdot 2=4$$ Ultimately, what "matters" is the reciprocal nature of the opposite half-angle tangents, which we get immediately from the cyclic nature of the quadrilateral. Your (neat!) formulas for those tangents confirm that reciprocal nature, but are themselves a bit like bread-saws with regard to the overall argument. :) ... Cheers! – Blue Jun 20 '22 at 13:05
  • @Blue: How $(x+\frac{1}{x})\geq2$? I do not see the standart version of the $AM-GM$ inequality here. Sorry, I couldn't sleep last night. – Emmanuel José García Jun 20 '22 at 13:33
  • I mean, what version of the $AM-GM$ inequality are you using here? – Emmanuel José García Jun 20 '22 at 13:43
  • For example, in Z Ahmed's answer, I can perfectly see what version of $AM-GM$ he is using. But my objection is that to prove the version he uses we need more sophisticated means than the ones I use. – Emmanuel José García Jun 20 '22 at 14:02
  • 1
    The relation $x+\frac1x\geq2$ is an application of the AM-GM instance $a+b\geq 2\sqrt{ab}$. See the picture-proof at the top of the Wikipedia "Inequality of arithmetic and geometric means" entry, or the variant at my Trigonography site. – Blue Jun 20 '22 at 17:27
  • That is definitely much simpler. Perhaps asking about the following inequality is less obvious: $$\sum_{cyc}\tan\frac\alpha2\tan\frac\beta2>\frac{a-c}{s-a}-\frac{a-c}{s-c}+\frac{b-d}{s-b}-\frac{ b-d}{s-d}$$. – Emmanuel José García Jun 21 '22 at 12:23

1 Answers1

1

Use AM_GM

$F=\frac{s-a}{s-c}+\frac{s-b}{s-d}+\frac{s-c}{s-a}+\frac{s-d}{s-b}\ge 4 \sqrt[4]{\frac{s-a}{s-c}\frac{s-b}{s-d}\frac{s-c}{s-a}\frac{s-d}{s-b}}=4.$

Z Ahmed
  • 43,235