Idk2018 wrote: Given convex hexagon $ABCDEF$, inscribed in the circle. Prove that $AC*BD*CE*EA*FB \geq 27 AB * BC * CD * DE * EF * FA$ How much is this question?

Feridimo wrote: Idk2018 wrote: Given convex hexagon $ABCDEF$, inscribed in the circle. Prove that $AC*BD*CE*EA*FB \geq 27 AB * BC * CD * DE * EF * FA$ How much is this question? P3

Tuleuchina wrote: Feridimo wrote: Idk2018 wrote: Given convex hexagon $ABCDEF$, inscribed in the circle. Prove that $AC*BD*CE*EA*FB \geq 27 AB * BC * CD * DE * EF * FA$ How much is this question? P3 Can you write on other questions?(izho 2020)

Idk2018 wrote: Given convex hexagon $ABCDEF$, inscribed in the circle. Prove that $AC*BD*CE*EA*FB \geq 27 AB * BC * CD * DE * EF * FA$ Probably $DF$ should also appear in the LHS? Otherwise the RHS is of higher degree with respect to the circumradius $R$ and thus for large $R$ the inequality will be false.

Some say inversion helps, but generally pizdets

Where did you get the problem from? It doesn't seem to be available on the official website. I wanted the problems of IZhO 2020 of other subjects...

Let $$X = AC\cdot BD\cdot CE\cdot DF\cdot EA\cdot FB \quad \text{and}\quad Y=AB \cdot BC\cdot CD\cdot DE \cdot EF\cdot FA$$Using Ptolomy's theorem in quadrilateral $ABCD$ we get $$AC \cdot BD = AB\cdot CD + BC \cdot AD = AB\cdot CD + \frac{BC \cdot AD}{2} + \frac{BC \cdot AD}{2} \geq 3\sqrt[3]{\frac{AB \cdot BC^2\cdot CD\cdot AD^2}{4}} $$We can obtain similar inequalities by applying Ptolomy's theorem in quadrilaterals in $BCDE$, $CDEF$, $DEFA$, $EFAB$, $FABC$. Multiplying these six inequalities side by side we get $$X \geq \frac{27}{4}\cdot \left(Y\cdot AD \cdot BE \cdot CF\right)^{2/3}$$Next, by using Ptolomy's theorem for quadrilateral $ABCE$ we get $$AC \cdot BE = AB\cdot CE + BC \cdot EA \geq 2\sqrt{AB \cdot BC\cdot CE\cdot EA} $$Writing the same inequality for quadrilaterals $BCDF$, $CDEA$, $DEFB$, $EFAC$, $FABD$ and multiplying the resulting inequalities side by side we obtain $$AD \cdot BE \cdot CF \geq 8 \cdot Y^{1/2}$$So $$X \geq \frac{27}{4}\cdot \left(Y\cdot AD \cdot BE \cdot CF\right)^{2/3} \geq \frac{27}{4}\cdot \left(Y\cdot 8 \cdot Y^{1/2}\right)^{2/3} = 27\cdot Y$$

$\text{Geometric inequality+cyclic polygon+Sides }\implies \text{Ptolemy}$

By inversion at $A$, this is equivalent to \[ (a+b)(b+c)(c+d)(a+b+c+d) \ge 27abcd \]which follows from AM-GM on the four factors when writing $a = \tfrac 12a + \tfrac 12a$ and $d = \tfrac 12d + \tfrac 12d$.

Let the circumcircle of $ABCDEF$ be $\Gamma$ and $\psi(ABCDEF)=\frac{AB\cdot BC\cdot CD\cdot DE\cdot EF\cdot FA}{AC\cdot CE\cdot EA\cdot BD\cdot DF\cdot FB}$. We wish to show that $\psi\le\frac{1}{27}$ for all hexagons. First, note that $\psi$ can be written in terms of cross-ratios like $\frac{AB\cdot CD}{AC\cdot BD}=|(BC,AD)|$, so $\psi$ is preserved under projective transformations. Since the space of convex hexagons inscribed in $\Gamma$ is compact (we allow consecutive vertices to coincide, and define $\psi$ to be $0$ in this case even if the denominator is $0$) and $\psi$ is continuos, there is a hexagon that maximizes $\psi$. Let $P=FB\cap CE$. As we change $A$ on the ark $FB$, $\frac{AB\cdot FA}{AC\cdot EA}\propto\frac{S(ABF)}{S(ACE)}\propto\frac{\text{dist}(A,BF)}{\text{dist}(A,CE)}\propto\frac{\sin BPA}{\sin CPA}$, and is maximized when the ray $PA$ is "farthest" from rays $PB$ and $PC$, so $PA$ has to be tangent to $\Gamma$, and, similarly, $PD$ is tangent to $\Gamma$. Since $AD$ is the polar of $P$ with respect to $\Gamma$, $AD$, $BE$ and $CF$ concur at $O$. Take a projective transformation preserving $\Gamma$ and taking $O$ to the center of $\Gamma$. Since the tangent to $\Gamma$ at $A$ is now parallel to $FB$ ($P$ goes to infinity), $FA=AB$, and, in a similar manner, all sides of $ABCDEF$ are equal, so $\psi(ABCDEF)=\frac{1}{27}$, as desired.

Denote the inequality $L \ge 27 R$ and denote the product $AD*BE*CF=M$. From ptolemy on $ABCD$ we get that $AC*BD=BC*AD+AB*CD$. Derive six equalities from $ABCD,BCDE,CDEF,DEFA,EFAB,FABC$ and multiply to get that $$L^2=(BC*AD+AB*CD)(CD*BE+BC*DE)...etc$$and from cauchy inequality get that: $$L^2\ge(R^{1/6}M^{2/6}+R^{1/3})^6 (\star)$$From ptolemy on $ACDF$ : $AD*CF=CD*AF+CA*DF $. Derive the same inequalities from $BDEA,CEFB$ and repeat the same cauchy -schwarz story as before to get that: $$M^2 \ge (R^{1/3}+L^{1/3})^3$$Use this on $(\star)$ to finaly get that $$L^{1/3} \ge R^{1/6}(R^{1/3}+L^{1/3})^{1/2}+R^{1/3}$$multiply both sides with $R^{1/6}(R^{1/3}+L^{1/3})^{1/2}-R^{1/3}$ and $$(R^{1/6}(R^{1/3}+L^{1/3})^{1/2}-R^{1/3} )L^{1/3} \ge R^{1/3} L^{1/3}$$$$(R^{1/3}+L^{1/3})^{1/2} \ge 2R^{1/6}$$$$L^{1/3} \ge 3R^{1/3}$$

Idk2018 wrote: Given convex hexagon $ABCDEF$, inscribed in the circle. Prove that $AC\cdot BD\cdot CE\cdot DF\cdot EA\cdot FB \geq 27\cdot AB \cdot BC\cdot CD\cdot DE\cdot EF\cdot FA$

Functional_equation wrote: Idk2018 wrote: Given convex hexagon $ABCDEF$, inscribed in the circle. Prove that $AC\cdot BD\cdot CE\cdot DF\cdot EA\cdot FB \geq 27\cdot AB \cdot BC\cdot CD\cdot DE\cdot EF\cdot FA$

This is just Ptolomy theorem as almost all ! the above solutions

Kamran011 wrote: This is just Ptolomy theorem as almost all ! the above solutions I have just learned Inversion . For Pratica, I use Inversion, not Ptolemy, in such problems.

The discussion is over. Please do not write anything extra. Because the forum is spoiled. (I will ask the mods to delete these meaningless posts)

extra note: i think it's okay if a solution is repeated, and frankly i don't see why people get so upset over repeated solutions in threads. There are multiple reasons as to why someone may post their solution other than posting something new (for storage, lets ppl see author's intended solution if they put it in a handout, etc.). In addition, I think the accusation of "upvote farming" here is a bit ridiculous - at least they are posting something that may be socially productive, and even if they get upvotes for this 1. upvotes do not matter, and 2. their post is of value anyways. So let the solution stay there and don't worry about upvote farming.

if we invert to a circle whose center is at point A, and after simple calculations come to prove the following expression. $(a+b)(b+c)(c+d)(a+b+c+d)\ge27abcd$ which is simply derived from Cauchy inequality.

Invert around $A$. By inversion distance formula spam, what we want to show becomes $$\frac{B^{\ast}D^{\ast}\cdot C^{\ast}E^{\ast}\cdot D^{\ast}F^{\ast}\cdot F^{\ast}B^{\ast}}{B^{\ast}C^{\ast}\cdot C^{\ast}D^{\ast}\cdot D^{\ast}E^{\ast}\cdot E^{\ast}F^{\ast}}\geq27$$Now, note that $B^{\ast},C^{\ast},D^{\ast},E^{\ast},F^{\ast}$ lie on a line in that order. Thus, let $B^{\ast}C^{\ast}=x,C^{\ast}D^{\ast}=y,D^{\ast}E^{\ast}=z,E^{\ast}F^{\ast}=w$. Then, what we want to show just becomes $$(x+y)(y+z)(z+w)(w+x+y+z)\geq 27wxyz.$$Note that in the original problem, a regular hexagon is an equality case. Then, after inverting a regular hexagon, we would have $(x:y:z:w)=(2:1:1:2)$ from 30-60-90 triangles. Hence, $(x:y:z:w)=(2:1:1:2)$ is an equality case. This will motivate the following steps. By AM-GM on $\frac{3}{2}x$ and $3y$ with weights $2/3$ and $1/3$ respectively, $$x+y\geq (\frac{3}{2}x)^{2/3}(3y)^{1/3}.$$ By AM-GM on $y$ and $z$, $$y+z\geq 2\sqrt{yz}.$$ By AM-GM on $3z$ and $\frac{3}{2}w$ with weights $1/3$ and $2/3$ respectively, $$z+w\geq (\frac{3}{2}w)^{2/3}(3z)^{1/3}$$ Finally, by AM-GM on $3x,6y,6z,3w$ with weights $1/3,1/6,1/6,1/3$ respectively, we have $$w+x+y+z\geq (3x)^{1/3}(6y)^{1/6}(6z)^{1/6}(3w)^{1/3}.$$ Note that the coefficents and weights were chosen so that $(2:1:1:2)$ is the equality case. Multiplying everything together, $$(x+y)(y+z)(z+w)(w+x+y+z)$$$$\geq (\frac{3}{2}x)^{2/3}(3y)^{1/3}(2\sqrt{yz})((\frac{3}{2}w)^{2/3}(3z)^{1/3})(3x)^{1/3}(6y)^{1/6}(6z)^{1/6}(3w)^{1/3}=27wxyz,$$as desired.

Inverting at $A$ and writing all lengths in terms of the new diagram, the inequality simplifies to $$\frac{B^*D^*\cdot C^*E^*\cdot D^*F^*\cdot B^*F^*}{B^*C^*\cdot C^*D^*\cdot D^*E^*\cdot E^*F^*}\ge 27$$ Letting $a = B^*C^*$, $b=C^*D^*$, $c=D^*E^*$, $d=E^*F^*$ we have by AM-GM $$(a+b)(b+c)(c+d)(a+b+c+d)\ge (3\cdot 2^{-\frac{2}{3}}a^{\frac{2}{3}}b^{\frac{1}{3}})(2b^{\frac{1}{2}}c^{\frac{1}{2}})(3\cdot 2^{-\frac{2}{3}}c^{\frac{2}{3}}d^{\frac{1}{3}})(6\cdot 2^{-\frac{2}{3}}a^{\frac{1}{3}}b^{\frac{1}{6}}c^{\frac{1}{6}}d^{\frac{1}{3}})=27abcd$$ as desired.

Invert about $A$. By inversion distance formula, it suffices to show that $$\frac{BD \cdot CE \cdot DF \cdot FB}{BC \cdot CD \cdot DE \cdot EF} \geq 27.$$As $B, C, D, E, F$ are collinear in this order, it suffices to show the inequality $$(a+b)(b+c)(c+d)(a+b+c+d) \geq 27abcd.$$This is straightforward by weighted AM-GM with the equality case $(2k, k, k, 2k)$.

Invert about $A.$ Simplifying with the inversion distance formula, we get that we need to prove that $BD \cdot DF \cdot FB \cdot CE \ge 27 BC \cdot CD \cdot DE \cdot EF,$ for points $B,C,D,E,F$ lying on a line in that order. This is equivalent to $(a+b)(b+c)(c+d)(a+b+c+d) \ge 27abcd$ for reals $a,b,c,d.$ This is proven by multiplying the four inequalities $\frac{a+b}{3} \ge \frac{a^{\frac{2}{3}}b^{\frac{1}{3}}}{2^{\frac{2}{3}}}, \frac{b+c}{2} \ge \sqrt{bc}, \frac{c+d}{3} \ge \frac{c^{\frac{1}{3}}d^{\frac{2}{3}}}{2^{\frac{2}{3}}},$ and $\frac{a+b+c+d}{6} \ge \frac{a^{\frac{1}{3}}b^{\frac{1}{6}}c^{\frac{1}{6}}d^{\frac{1}{3}}}{2^{\frac{2}{3}}},$ obtained from AM-GM.

We invert about $A$ with radius $1$. By the inversion distance formula (and a lot of cancelation), we get the problem reduced to proving that \[\frac{B^*C^* + C^*D^* + D^*E^* + E^*F^*)(B^*C^* + C^*D^*)(C^*D^* + D^*E^*)(D^*E^* + E^*F^*)}{B^*C^* \cdot C^*D^* \cdot D^*E^* \cdot E^*F^*} \geq 27\]Notice that these are all free variables so we let $B^*C^* = b$ and cyclically the same. We use the following four facts to finish this inequality. \[\frac{b}{2} + \frac{b}{2} +c+d+\frac{e}{2} + \frac{e}{2} \geq 6 \sqrt[6]{\frac{b}{2} \cdot \frac{b}{2} \cdot c \cdot d \cdot \frac{e}{2} \cdot \frac{e}{2}}\]\[\frac{b}{2} + \frac{b}{2} + c \geq 3 \sqrt[3]{\frac{b}{2} \cdot \frac{b}{2} \cdot c}\]\[c+d \geq 2 \sqrt{c+d}\]\[d+\frac{e}{2} + \frac{e}{2} \geq 3\sqrt[3]{d \cdot \frac{e}{2} \cdot \frac{e}{2}}\] All of these follow from AM-GM and when multiplied obviously produce the result.