Quyidagi masala IZHO 2020 dan 'salom covid' degan masala; har safar ushbu masalani eslaganda, covid oldidagi sukunat esga tushadi. Lekin, juda ajoyib masala bo'lgan, oltiburchaklar uchun Ptolemey tengsizligiga o'xshash analog vazifasini bajarishi mumkin (adashmasam, bu ham Sedrakyan masalasi).
1-yechim. (proyektiv yondashuv) Proyektiv ma'noda biz uchun aylana ham to'g'ri chiziqdir (ikkalasi ham $\mathbb{RP}^1$ ga homeomorf).
Aytaylik, bizga biror $\ell$ (proyektiv) to'g'ri chiziqdagi $B,C,D,E,F$ (berilgan tartibdagi yotuvchi) nuqtalar berilgan bo'lsin va $A=\equiv \infty$ nuqtani ham alohida olamiz (2-chizma ga qarang). Ushbu nuqtalarni aylanaga ichki chizilgan oltiburchak hosil qiladi deb hisoblab, masalada berilgan tengsizlikni ko'rsatamiz.
 |
| 2-chizma |
$BC=x$, $CD=y$, $DE=z$, $EF=t$ deb olaylik, u holda bizga berilgan tengsizlik \[ (x+y)(y+z)(z+t)(x+y+z+t) \ge 27 xyzt \] ga ekvivalent bo'ladi, chunki $\frac{AB\cdot FA}{AC\cdot EA} = \frac{\overline{\infty B} \cdot \overline{F \infty}}{\overline{\infty C}\cdot \overline{E \infty}} = 1$. Yuqoridagi tengsizlikni tekshirish qiyin emas, $\frac{x}{2}=y=z=\frac{t}{2}$ tenglik holini saqlagan holda Koshi tengsizligini qo'llash mumkin: \[ x+y = \frac{x}{2}+\frac{x}{2}+y \ge 3\sqrt[3]{\frac{x^2y}{4}}, \] \[ y+z \ge 2\sqrt{yz}, \] \[ z+t = z+\frac{t}{2}+\frac{t}{2}\ge 3\sqrt[3]{\frac{zt^2}{4}} ,\] \[ x+y+z+t \ge \frac{x}{2}+\frac{x}{2}+y+z+\frac{t}{2}+\frac{t}{2}\ge 6\sqrt[6]{\frac{x^2yzt^2}{4^2}}. \] Yuqoridagi tengsizliklarni ko'paytirish orqali bizga kerakli tengsizlikni olamiz.
Endi, aylanaga ichki chizilgan $ABCDEF$ oltiburchakka qaytadigan bo'lsak, $A$ markazli inversiya qo'llash orqali tashqi aylanani biror $\ell$ t/ch ga, $A$ ni $\infty$ ga va $B,C,D,E,F$ nuqtalarni $\ell$ t/ch dagi $B',C',D',E',F'$ nuqtalarga o'tkazadi (1-chizma ga qarang). Albatta, inversiyaning asosiy xossalaridan biri bu masofalar orasidagi nisbat edi, yani bizda \[ X'Y' =XY\cdot \frac{R^2}{AX\cdot AY}. \] Ushbu nisbatlardan foydalansak, $ABCDEF$ oltiburchak uchun berilgan tengsizlikni $\ell$ t/ch dagi $\infty,B',C',D',E',F'$ nuqtalar uchun biz yuqorida isbotlangan tengsizlikka olib kelishimiz mumkin.
2-yechim. Ptolemey teoremasidan foydalanamiz. Aylanaga ichki chizilgan $ABCDEF$ oltiburchakda bir nechta siklik to`rtburchaklar bor. Oldiniga $AD$ atrofidagi $ABDF$, $ACDE$, $ABCD$, $AFED$ siklik to`rtburchaklarni olamiz va ular uchun Ptolemey teoremasini qo'llab quyidagilarni yozamiz: \[ ABDF : \, \, \, \, \, \, \, \, AB\cdot DF+FA\cdot BD = AD\cdot FB, \] \[ ACDE : \, \, \, \, \, \, \, \, CD\cdot EA + DE\cdot AC = AD\cdot CE, \] \[ ABCD : \, \, \, \, \, \, \, \, AB\cdot CD + BC\cdot AD = AC\cdot BD, \] \[ AFED : \, \, \, \, \, \, \, \, FA\cdot DE + FE\cdot AD = EA\cdot DF. \] Agar yuqoridagilarning har biriga Koshi tengsizligini qo'llaydigan bo'lsak, mos ravishda \[ 2\sqrt{AB\cdot DF\cdot FA\cdot BD} \le AD\cdot FB, \] \[ 2\sqrt{CD\cdot EA \cdot DE\cdot AC} \le AD\cdot CE, \] \[ 3\sqrt[3]{AB\cdot CD \cdot \frac{BC^2\cdot AD^2}{4}} \le AB\cdot CD + \frac{BC\cdot AD}{2} + \frac{BC\cdot AD}{2} = AC\cdot BD, \] \[ 3\sqrt[3]{FA\cdot DE \cdot \frac{FE^2\cdot AD^2}{4}} \le FA\cdot DE + \frac{FE\cdot AD}{2} + \frac{FE\cdot AD}{2} = EA\cdot DF \] larni topamiz. Yuqoridagi birinchi ikkita tengsizlikni ko'paytirib kvadratga oshirsak, \[ 16 (AB\cdot FA \cdot CD\cdot DE) (AC\cdot EA\cdot BD\cdot DF) \le AD^4 \cdot FB^2\cdot CE^2 \quad (*)\] keladi. Keyingi imita tengsizlikni kubga oshirish va ko'paytirish orqali esa \[ \frac{27^2}{16} (AB\cdot BC\cdot CD\cdot DE \cdot EF\cdot FA) (BC\cdot EF)\cdot AD^4 \le (AC\cdot EA\cdot BD\cdot DF)^3 \quad (**) \] ni hosil qilamiz.
Demak, $(*)$ va $(**)$ tengsizliklarni ko'paytirish natijasida bizga kerakli bo'lgan tengsizlikni keltirib chiqarishimiz mumkin.