Suppose, $ P,Q,R,S,T $ are the midpoints of $ AB,BC,CD,DE,EA $. Easy to see that, $ S_{A'B'C'D'E'}=4S_{PQRST} $. So the inequality turns out to $ 2S_{PQRST}\ge S_{ABCDE}\ge \frac {4S_{PQRST}}{3} $. We will prove this for any convex pentagon. Part 1:- $ 2S_{PQRST}\ge S_{ABCDE} $ Proof:- At first we will show that, among the vertices $ A,B,C,D,E $, there exists one (call it $ A $ for now), such that $ AB\cap CD $ and $ D $ are on opposite sides of $ C $ and also $ AE\cap CD $ and $ C $ are on opposite sides of $ D $. To prove thia consider the contradiction. In that case, among the pairs $ (\angle A,\angle B),(\angle B,\angle C),... $, there will be $ 3 $ pairs with sum $ <180 $. But that is not possible, because among the $ 3 $ pairs there will be $ 2 $ with no common vertex and in that case the angle left will be $ >180 $, which is not possible. So WLOG, suppose, it holds for $ A $. Then note that, the inequality is equivalent to $ S_{PRT}\ge S_{PQB}+S_{TSE} $. $ X,Y $ be the midpoints of $ BR,RE $. clearly, $ S_{PXB}\ge S_{PQB} $ and similar for the other. So its enough to prove that, $ S_{PRT}\ge S_{PXB}+S_{TYE}\iff S_{APT}\ge 0 $. So done. Part 2:- $ 3S_{ABCDE}\ge 4S_{PQRST} $. Proof:- Suppose, $ A_1=BD\cap CE $ and similarly define $ B_1,C_1,D_1,E_1 $. Consider the triangles, $ \Delta ABE, \Delta AED $ and other such 3 triangles. WLOG, suppose, $ \Delta AEB $ has maximum area among them. The note that, $ EC\cap AB $ and $ A $ are on opposite sides of $ B $ and similar for $ E $. So $ S_{A_1EB}\le S_{AEB} $. So $ 4S_{AEB}\ge S_{AEB}+S_{A_1EB}+S_{BDC}+S_{CED}\ge S_{ABCDE} $. So $ S_{A_1B_1C_1D_1E_1}\le \frac {3S_{ABCDE}}{4} $. Suppose, $ f(ABCDE)=3S_{ABCDE}-4S_{PQRST}$ $=\sum_{similarly} S_{AEB}-ABCDE=\sum_{similarly} S_{AD_1B}-S_{A_1B_1C_1D_1E_1} $. Now note that, $ S_{AC_1B}\ge S_{AE_1C_1}\implies S_{AD_1B}\ge S_{C_1D_1E_1} $. So $ f(ABCDE)\ge f(A_1B_1C_1D_1E_1) $. Similarly for $ A_1B_1C_1D_1E_1 $ define $ A_2B_2C_2D_2E_2 $. So $ f(ABCDE)\ge \lim_{n\to \infty} f(A_nB_nC_nD_nE_n)=0 $ since the area of $ A_nB_nC_nD_nE_n $ is exponentially decreasing. So done.

This one is really really much easier if we take coordinate system. Let $O$ be the origin. Easy to see that $A'$ is orthocenter of $ABC$, so $x_{A'}=x_A+x_b+x_C$ and analogously. So now we can compute their surfaces and it turns out to be $S_{A'B'C'D'E'}=2S_{ABCDE}+\sum S_{AOC}$. But note that maybe we should count $S_{AOC}$ as negative, if $\angle CBA<90^{o}$. But $S_{AOC}=S_{AOB}+S_{BOC}-S_{ABC}$, so $S_{A'B'C'D'E'}=2S_{ABCDE}+\sum S_{AOC}=4S_{ABCDE}-\sum S_{ABC}$. It's obvious that $\sum S_{ABC} < 2S_{ABCDE}$ and it can be proven that some four out of these five triangles have greater area than $S_{ABCDE}$, so we're done. Last stament may be pretty hard, but I will leave it as an exercise ; p. To obtain key equality, which I don't know how to prove in standard way, you can use as well complex numbers, it's practically the same calculus. And believe me, I'm really against coordinate system and complex numbers, but they really help in this problem.

goodar2006 wrote: The pentagon $ABCDE$ is inscirbed in a circle $w$. Suppose that $w_a,w_b,w_c,w_d,w_e$ are reflections of $w$ with respect to sides $AB,BC,CD,DE,EA$ respectively. Let $A'$ be the second intersection point of $w_a,w_e$ and define $B',C',D',E'$ similarly. Prove that \[2\le \frac{S_{A'B'C'D'E'}}{S_{ABCDE}}\le 3,\]where $S_X$ denotes the surface of figure $X$. Proposed by Morteza Saghafian, Ali khezeli If $O, O_a, O_b$ are circumcenters of $w$, $w_a$, $w_b$, respectively, then $OAO_aB, OBO_bC, O_aBO_bB'$ are all parallelograms. So, if we assign complex numbers $a, b, c, d, e$ to $A, B, C, D, E$ and $0$ to $O$, then we see that $O_a = a+b$ and $O_b = b+c$. Thus we get $B' = a + b + c, C' = b+c+d$ and $$ A + C' = a + b + c + d = B' + D$$This means $AB'C'D$ is a parallelogram. In particular, $B'C' \parallel AD$ and $B'C' = AD$. So if $A'', B'', C'', D'', E'' $ are midpoints of $AB, BC, CD, DE, EA$, then $A''B''C''D''E''$ is similar to $A'B'C'D'E'$ with similarity ratio of $2$. Hence $$S_{A'B'C'D'E'} = 4 S_{A''B''C''D''E''}$$and we need to prove that $$\frac{1}{2} \leq \frac{S_{A''B''C''D''E''}}{S_{ABCDE}}\leq \frac{3}{4}$$If $$S = S_{AA''E''} + S_{BB''A''} + S_{CC''B''} + S_{DD''C''} + S_{EE''D''}$$then this inequality can be written as $$2S \leq S_{ABCDE}\leq 4S$$Without loss of generality assume either $EB \parallel DC$ or rays $EB$ and $DC$ intersect. Note that if ray $XY$ and line $MN$ intersect or are parallel, then $S_{YMN} \leq S_{XMN}$ since altitude from $X$ to $MN$ is bigger than the altitude from $Y$. Equality case is achieved if and only if $XY \parallel MN$. We will repeatedly use this fact to move points along the lines. First let's prove that $$S_{ABCD} \leq 4S$$Draw lines $AC$ and $AD$ and note that $S_{ABC} = 4S_{BB''A''}$ and $S_{AED} = 4S_{EE''D''}$. So we just need to prove that $$S_{ADC} \leq 4S_{AA''E''} + 4S_{CC''B''} + 4S_{DD''C''} = S_{AEB} + S_{BDC} + S_{EDC}$$Draw a line $\ell$ through $B$ parallel to $DC$. If $\ell$ doesn't intersect segment $AC$, then ray $BA$ intersects line $CD$. So $S_{ADC} \leq S_{BDC}$ and we are done. Otherwise, let $\ell$ intersect segment $AC$ at $B_1$ and note that $S_{B_1DC} = S_{BDC}$. If ray $EA$ intersects or parallel to the line $CD$, then $S_{ADC} \leq S_{EDC}$ and we are done again. Otherwise, ray $AE$ intersects line $CD$. Since $BB_1 \parallel CD$ and $BB_1$ intersects segment $AC$ we deduce that ray $AE$ also intersects line $BB_1$ and $S_{AB_1E} \leq S_{ABE}$. So if we move $B$ to $B_1$, then the left hand side of our inequality stays fixed while the right hand side is decreasing. Observe that rays $EB_1$ and $DC$ still intersect. So, if segments $EB_1$ and $AD$ intersect at $E_1$, then $S_{E_1DC} \leq S_{EDC}$ and $S_{AE_1B} \leq S_{AEB}$. So if we move $E$ to $E_1$, then we further decrease the right hand side while the left stays same. So we need to show that $$S_{ADC} \leq S_{AE_1B_1} + S_{B_1DC} + S_{E_1DC}$$which is equivalent to $$S_{E_1B_1C} \leq S_{B_1DC}$$But this is true because rays $DE_1$ and $CB_1$ intersect at $A$. Now, let's prove that $$2S \leq S_{ABCD}$$Multiplying both sides by $2$ and using ideas as abode we can write this inequality as $$S_{AEB} + S_{BDC} + S_{EDC} \leq S_{ABCD} + S_{ADC}$$If $B_1$ and $E_1$ are as above, then again we have $S_{BDC} = S_{B_1DC}$. Let $X$ be the intersection of $AC$ and $EB$, and let $Y$ be the intersection of $AD$ and $EC$. Then \begin{align*} S_{AEB} - S_{AEB_1} = S_{ABX} - S_{EB_1X} \leq S_{ABX} \leq S_{ABC} \end{align*}This means when we move $B$ to $B_1$, then the right hand side of our inequality decreases more than the left hand side. So we just need to prove our inequality for $B_1$. Next, observe that \begin{align*} S_{AEB_1} + S_{EDC} - (S_{AE_1B_1} + S_{E_1DC}) &= S_{EE_1A} + S_{EYD} - S_{E_1YC} \leq S_{EE_1A} + S_{EYD}\leq S_{AED} \end{align*}So, if we move $E$ to $E_1$, then the right hand side again will decrease more than the left hand side. This means we just need to prove $$S_{AE_1B_1} + S_{B_1DC} + S_{E_1DC} \leq 2S_{ADC}$$This follows easily from $$S_{AE_1B_1} + S_{B_1CD} \leq S_{ADC}$$and $$S_{E_1DC} \leq S_{ADC}$$