Zhautykov Olympiad 2015 problem 6 This links discusses the olympiad problem which none of students could solve , meaning it is very hard.
Question:
The area of a convex pentagon $ABCDE$ is $S$, and the circumradii of the triangles $ABC$, $BCD$, $CDE$, $DEA$, $EAB$ are $R_1$, $R_2$, $R_3$, $R_4$, $R_5$. Prove the inequality
$$R_1^4+R_2^4+R_3^4+R_4^4+R_5^4\geq {4\over 5\sin^2 108^\circ}S^2$$
A month ago,I tried to solve this problem,because I know the following Möbius theorem(1880):
Theorem:
Let $a,b,c,d,e$ are area of $\Delta ABC,\Delta BCD,\Delta CDE,\Delta DEA,\Delta EAB$,then we have
$$S^2-S(a+b+c+d+e)+ab+bc+cd+de+ea=0$$
This is a result by Gauss 1880,you can see this paper (for proof):(Gauss Carl Fridrich Gauss Werke Vol 4.2ter Abdruck,1880:406-407)
Using this Theorem ,in 2002 ,Chen Ji and Xiongbin proved this result:
$$a^2+b^2+c^2+d^2+e^2\ge\dfrac{20S^2}{(5+\sqrt{5})^2}$$
I use the well known result: $$a\le\dfrac{3\sqrt{3}}{4}R^2_{1},b\le\dfrac{3\sqrt{3}}{4}R^2_{2},c\le\dfrac{3\sqrt{3}}{4}R^2_{3},d\le\dfrac{3\sqrt{3}}{4}R^2_{4},e\le\dfrac{3\sqrt{3}}{4}R^2_{5}$$
so we have
$$R_1^4+R_2^4+R_3^4+R_4^4+R_5^4\ge\dfrac{16}{27}(a^2+b^2+c^2+d^2+e^2)
\ge\dfrac{16\cdot 20}{27(5+\sqrt{5})^2}S^2$$
But I found
$$\dfrac{16\cdot 20}{27(5+\sqrt{5})^2}<\dfrac{32}{5(5+\sqrt{5})}={4\over 5\sin^2 108^\circ}$$
so my work can't solve this contest problem. If we can solve following question,then the Olympiad problem can be solved:
Question 2:
$$a^2+b^2+c^2+d^2+e^2\ge\dfrac{54}{5(5+\sqrt{5})}S^2$$
Best Answer
Dan Schwarz (one of the major problem proposers for EGMO, RMM, Balkan...) has posted a solution at here. I'll briefly sketch it here.
First, one takes the midpoints $M_1$, ..., $M_5$ of $A_1A_2$, ..., $A_5A_1$. Then at each angle $A_i$, one takes the circumcircle of triangle $M_{i-1}A_iM_{i+1}$ (which has radius $\frac 12 R_i$) and the point diametrically opposite $A_i$ on it, say $X_i$. This gives a quadrilateral $A_iA_{i-1}X_iA_{i+1}$, with area at most $(\frac 12 R_i)^2 \cdot 2\sin A_i$. Consider all five of these quadrilaterals, as shown.
It's not too hard to show that these five quadrilaterals cover the entire pentagon (look at perpendicular bisectors). So summing gives $$ S \le \frac{1}{2} \sum_i R_i^2 \sin A_i. $$ Then, by the Cauchy-Schwarz Inequality, we have $$ 4S^2 \le \left( \sum_i R_i^2 \sin A_i \right)^2 \le \left( \sum_i R_i^4 \right) \left( \sum_i \sin^2 A_i \right). $$ So it remains to show $$\sum_i \sin^2 A_i \le 5\sin^2 (108^{\circ})$$ reducing this to a purely algebraic problem (with $0^{\circ} \le A_i \le 180^{\circ}$ and $\sum A_i = 540^{\circ}$).
It's tempting to try and apply, say, Jensen's Inequality, but the function $\sin^2 x$ has a few inflection points, so one has to proceed more delicately using a fudging argument (along the lines of the $n-1$ equal-value trick). Indeed, we do this by showing that if $A_1 \le A_2 \le A_3 \le A_4 \le A_5$ and $A_1 < 108^{\circ}$, then we can replace $A_1$ with $108^{\circ}$ and $A_5$ by $A_1+A_5-108^{\circ}$ while increasing $\sum_i \sin^2 A_i$. Repeating this process until all $A_i$ are equal completes the proof.