Exactly, what "convex polygon that is symetric to some point" means?

I believe it means that there is a point $O$ such that for any vertex $X$ of the polygon, the reflection of $X$ across $O$ is also a vertex of the polygon

Nice problem! Take vertices $A,B$ of the polygon such that the area of $OAB$ is maximum over all possible pairs of vertices. Wlog we may consider that $OA\perp OB$ and $OA=OB$, since an afine transformation preserves the ratios between areas. Let $A',B'$ be the symmetric of $A,B$ with respect to $O$. Draw a rectangle with sides parallel to $AB,BA',A'B',B'A$ and circumscribed to the polygon, such that on each one of its sides there is at least one vertex of the polygon. The area of this rectangle is $(\ell+2u)(\ell+2v)$, where $\ell$ is the sidelength of $ABA'B'$, and $u,v$ are the distances from the sides of the rectangle respectively parallel to $AB,BA'$, and $AB,BA'$. This rectangle is the result of applying the initial afine transformation to a parallelogram since afine transformations preserve parallelism. Note that the area of the polygon is at least $\ell^2+\ell u+\ell v$, ie the area of $ABA'B'$, plus at least the area of four triangles, one of which has vertices $A,B$, and the vertex where the polygon touches the rectangle side parallel to $AB$, and the other three are analogously constructed on sides $BA',A'B',B'A$. Consider also the square whose sides are the perpendiculars to the diagonals $AA'$ and $BB'$, through $A$ and $A'$, and through $B$ and $B'$, respectively. Clearly this square contains the polygon inside, otherwise the area of $OAB$ is not maximum, and the square has area $2\ell^2$. It remains only to prove that it is impossible to have simultaneously $2\ell^2>\sqrt{2}\ell(\ell+u+v)$ and $(\ell+2u)(\ell+2v)>\sqrt{2}\ell(\ell+u+v)$. From the first inequality we find that $u+v<\ell(\sqrt{2}-1)$, and from the second one, $4uv>(\sqrt{2}-1)\ell(\ell-\sqrt{2}(u+v))$, and when inserting the first one into the second one, we obtain $4uv>(\sqrt{2}-1)^2\ell^2>(u+v)^2$, clearly false because of the AM-GM inequality. The conclusion follows, equality being reached simultaneously for the square and rectangle considered, and only if the original polygon is an octagon where, after the afine transformation, $u=v=\frac{\sqrt{2}-1}{2}\ell$.

April wrote: Let $P$ be a polygon that is convex and symmetric to some point $O$. Prove that for some parallelogram $R$ satisfying $P\subset R$ we have \[\frac{|R|}{|P|}\leq \sqrt 2\] where $|R|$ and $|P|$ denote the area of the sets $R$ and $P$, respectively. Is there any better bound? We can find that when P is an octagon it can be smaller than this bound.

I heard the best constant is $\frac{4}{3}$, but it's hard to prove.

timon92 wrote: I heard the best constant is $\frac{4}{3}$, but it's hard to prove. If so, then $\frac{4}{3}$ is the best bound. Consider the regular hexagon. Where do you hear this?

litongyang wrote: April wrote: Let $P$ be a polygon that is convex and symmetric to some point $O$. Prove that for some parallelogram $R$ satisfying $P\subset R$ we have \[\frac{|R|}{|P|}\leq \sqrt 2\] where $|R|$ and $|P|$ denote the area of the sets $R$ and $P$, respectively. Is there any better bound? We can find that when P is an octagon it can be smaller than this bound. In my solution, I meant equality between $\sqrt{2}$ times the area of the polygon, and the area of either of the two parallelograms considered. Clearly, picking other choices for parallelograms probably improves the bound. I do not know however if a smart parallelogram choice that can be worked out in "competition time" can improve the bound significantly, so I guess the problem itself as it is, is a good IMO problem, but finding the best bound is a more time-consuming task. The $\frac{4}{3}$ bound seems pretty good, if I have some free time I'll try to kick it around and see if something comes up.

litongyang wrote: timon92 wrote: I heard the best constant is $\frac{4}{3}$, but it's hard to prove. If so, then $\frac{4}{3}$ is the best bound. Consider the regular hexagon. Where do you hear this? At camp. One of the teacher said this, but he didn't prove this.

It is a very nice problem. First, some notation. For a point to be mentioned in this post, it must be an element of P. For any point or line $x$, denote $x'$ as the reflection of $x$ through $O$. Also, if $l$ is a support line of P, then $l'$ is the other support line parallel to $l$. For any two lines $l$ and $m$, denote $lml'm'$ as the parallelogram bounded by the four lines. Finally, let $c$ be the minimum value of all parallelograms which cover $P$. Take any support lines $l$ and $l'$, and $A$ on $l$, $A'$ on $l'$. Then if $m$ is a support line parallel to $AA'$, and $B$ is on $m$, then $|ABA'B'| = |lml'm'| /2 \ge c/2$. Now consider the value of $|AXA'X'|$ for a point $X$, traveling on a path from $A$ to $B$ inside $P$. That value ranges from zero to greater than or equal to $c/2$, so there must exist some point $C$ so that $|ACA'C'| = c/2$. Consider support lines $n$ and $p$ parallel to $AC$ and $A'C$, and points $R$ and $S$ on $n$ and $p$ respectively. If the ratios of the correspondingly parallel sides of $npn'p'$ to the sides of $ACA'C'$ are $a$ and $b$ respectively, then $|npn'p'| = abc/2 \ge c$, so $ab \ge 2$. Also $|ARCSA'R'C'S'|=(a+b)c/4$. (Take the sum of $|ACA'C'|$ and the four triangles facing outward from the sides.) So $|P| \ge (a+b)c/4 \ge c\sqrt{ab}/2 \ge c/\sqrt{2}$.

We will construct two parallelograms $P_1\supseteq P$ and $P_2\supseteq P$ and prove at least one of them has area at most $\sqrt{2}|P|$. First, pick two vertices $A\neq B$ of $P$ such that triangle $OAB$ has the maximal area. Let $C$ and $D$ be reflections of $A$ and $B$ across $O$, respectively, and let $Q$ denote the parallelogram $ABCD$. Consider the lines through $A$ and $C$ parallel to $OB$, and the lines through $B$ and $D$ parallel to $OC$; these four lines form a parallelogram $P_1$. Let $P_2$ be the minimal parallelogram containing $P$ whose sides are all parallel to sides of $Q$. An example is drawn below, where the boundary of $P_1$ is colored red and the boundary of $P_2$ is colored blue: [asy][asy] size(7cm); pair A = (1.16,4.24); pair B = (-5.37,-0.8); pair A1 = (-0.68,3.77); pair A2 = (-4.38,1.45); pair A3 = (-3.09,-3.92); pair E = (-4.22,3.43); pair F = (6.53,5.04); pair G = (1.01,5.61); pair H = (-7.1,-0.65); draw(A--A1--A2--B--A3--(-A)--(-A1)--(-A2)--(-B)--(-A3)--cycle); draw(E--F--(-E)--(-F)--cycle,red); draw(G--H--(-G)--(-H)--cycle,blue); draw(A--(-A)); draw(B--(-B)); draw(A--B--(-A)--(-B)--cycle); draw(F--origin--E); dot((-3.04,2.48)); dot((4.05,3.13)); label("$A$",A,N); label("$B$",B,NW); label("$C$",(-A),S); label("$D$",(5.37,0.8), SE); label("$E$",E,NW); label("$F$",F,NE); label("$E'$",0.5*E,S); label("$F'$",0.5*F,S); label("$G$",(-3.04,2.48),N); label("$H$",(4.05,3.13),N); label("$G'$",A2,N); label("$H'$",(-A3),N); label("$O$",origin,SE); [/asy][/asy] Claim: $P\subseteq P_1$, and $|P_1| = 2|Q|$. Proof: Suppose some point $X\in P$ lies outside $P_1$; then either $S_{OXB} > S_{OAB}$ or $S_{OXA}>S_{OAB}$, contradiction. Because $A$ is the midpoint of side $EF$, it follows that $|P_1| = 2|Q|$. Next, we will bound $|P|$ in terms of $|Q|$. Suppose the boundary of $P_2$ touches polygon $P$ at points $G',H'$ (and their reflection about $O$) as labeled on the diagram; because $P$ is convex, $\triangle AG'B \subseteq (P\cap \measuredangle AOB)$ and $\triangle AH'D \subseteq (P\cap \measuredangle DOA)$ (the measured angles represent the set of points inside or on the boundary of that angle). Let $G,H$ be intersections of $\overline{OE},\overline{OF}$ with the boundary of $P_2$, and let $x=\frac{|OG|}{|OE'|}$, $y=\frac{|OH|}{|OF'|}$. Then \begin{align*} |P| &\ge 2(S_{AG'BO} + S_{AH'DO})\\ &= 2(S_{AGBO} + S_{AHDO})\\ &= \left(\frac{OG}{OE'}+\frac{OH}{OF'}\right)S_{ABD}\\ &= \frac{1}{2}(x+y)|Q|. \end{align*}Now we are almost done: observe that $|P_2| = xy|Q|$, so $$|P_1||P_2| = 2xy|Q|^2 \le \left(\frac{x+y}{\sqrt{2}}|Q|\right)^2 \le (\sqrt{2}|P|)^2$$so at least one of $|P_1|$ and $|P_2|$ is less than or equal to $\sqrt{2}|P|$.

Consider the parallelogram $R$ of minimal area which contains $P$. We shear/dilate the figure as a whole until $R$ is transformed to the unit square. Call the two heights of $R$ going through $O$ $h_1$ (vertical) and $h_2$ (horizontal). Now, if we consider the height to a pair of sides of $P$, $h'$, which is an angle $\alpha\ge45^o$ away from $h_1$, we know that the parallelogram induced by the lines perpendicular to $h_1$, $h'$ has to have area at least $1$. So, $\frac{\ell(h_1)\ell(h')}{\sin\alpha}\ge 1\implies \ell(h')\ge \sin\alpha$. So, the foot of $h'$ must lie on the exterior of the circle centered at the midpoint of $h_1$ and passing through $O$ (by polar coordinates). We get similar results for $\alpha< 45^o$, so the feet from $O$ to any side of the polygon must lie outside the region shown: [asy][asy] pair A,B,C,D; A=(1,0);B=(0,1);C=(-1,0);D=(0,-1); draw(A--B--C--D--A,dotted); draw(A--C);draw(B--D); label("$h_1$",(0,0.5),W); label("$h_2$",(0.5,0),S); draw(arc((0,0.5),0.5,0,180),red); draw(arc((-0.5,0),0.5,90,270),red); draw(arc((0,-0.5),0.5,180,360),red); draw(arc((0.5,0),0.5,-90,90),red); [/asy][/asy] However, as $h_1$,$h_2$ are the diameters of these circles, if the foot of an altitude is on the perimeter of this region, the extended side touches one of the vertices of the inner square - so it lies completely outside of the square. Hence, every edge of $P$ lies outside of the square, so $P$ contains this square. [asy][asy] pair A,B,C,D; A=(1,0);B=(0,1);C=(-1,0);D=(0,-1); draw(A--B--C--D--A,dotted); draw(A--C);draw(B--D); draw((0.7,0.7)--(0,0)--(-0.8,0.8),dashed); draw((1,1)--(-1,1)--(-1,-1)--(1,-1)--(1,1)); draw((1,0.4)--(0.4,1)); draw((-1,0.6)--(-0.6,1)); label("$x$",(0.4,0.4),SE); label("$y$",(-0.4,0.4),SW); dot((0.9,0.5)); dot((-0.7,0.9)); draw((-1,0)--(-0.7,0.9)--(0,1)--(0.9,0.5)--(1,0)); label("$h_1$",(0,0.5),W); label("$h_2$",(0.5,0),S); draw(arc((0,0.5),0.5,0,180),red); draw(arc((-0.5,0),0.5,90,270),red); draw(arc((0,-0.5),0.5,180,360),red); draw(arc((0.5,0),0.5,-90,90),red); [/asy][/asy] Now, consider the vertex of $P$ in quadrant I which is the furthest from $O$ in the $\frac{\pi}{4}$ direction, and call its distance along that direction $x$. Similarly, consider the distance to the furthest point along the $\frac{3\pi}{4}$ direction $y$. By minimality of $R$, we should have $xy\le \frac{1}{4}$. However, note that these vertices add at least the two triangles shown in the figure to $P$. Taking into account symmetry, we have that $$[P]\ge\frac{1}{2}+\frac{\sqrt{2}}{2}\left(x+y-\frac{\sqrt{2}}{2}\right)\ge \frac{1}{2}+\frac{\sqrt{2}}{2}\left(1-\frac{\sqrt{2}}{2}\right)=\frac{\sqrt{2}}{2}$$Hence, $\frac{|R|}{|P|}\le\sqrt{2}$, as desired.

Suppose that $\mathcal{R}$ is a parallelogram that contains $\mathcal P$, is symmetric across $O$, and has minimal area. It's clear that each side of $\mathcal R$ must contain at least one vertex. If there is a side that contains exactly one vertex $v$, then we perturb that side to be coincide with a side of $\mathcal P$ incident to $v$. It's easy to see that one of the two choices results in the smaller or same area. Therefore, we assume that each side of $\mathcal R$ must coincide with a side of $\mathcal P$. Take a suitable affine transformation so that $\mathcal{R}$ is a square. As shown in the figure below, let $\mathcal{R} = ABCD$ and $\overline{W_1W_2}$ be a side of $\mathcal P$ contained in $\overline{AB}$ ($A,W_1,W_2,B$ are in this order). We claim that the midpoint $W$ of $\overline{AB}$ must lie in $\overline{W_1W_2}$. Assume not; WLOG, let $W\in \overline{AW_1}$. We can see that pertubing the side $\overline{AB}$ to $\overline{A'B'}$ so that $W_1\in \overline{A'B'}$ results in $\mathcal{R}$ with a smaller area, contradiction. [asy][asy] size(6cm); defaultpen(fontsize(10pt)); pair A = (0,1); pair B = (1,1); pair C = (1,0); pair D = (0,0); pair W_1 = (0.6,1); pair W_2 = (0.75,1); pair W = (0.5,1); pair A_1 = (0,0.7); pair B_1 = extension(A_1,W_1,B,C); pair C_1 = (1,1)-A_1; pair D_1 = (1,1)-B_1; filldraw(A_1--W_1--W_2--(1,0.9)--C_1--(1,1)-W_1--(1,1)-W_2--(0,0.1)--D--cycle,yellow+opacity(0.5),dashed); draw(A--B--C--D--cycle,linewidth(1.5)); draw(A_1--B_1--C_1--D_1--cycle); dot('$A$',A,NW); dot('$B$',B,NE); dot('$C$',C,SE); dot('$D$',D,SW); dot('$W_1$',W_1,dir(280)); dot('$W_2$',W_2,S); dot('$W$',W,1.5*N); dot("$A'$",A_1,dir(180)); dot("$D'$",D_1,dir(180)); dot("$B'$",B_1,E); dot("$C'$",C_1,E); label("\Large$\mathcal{P}$",(0.5,0.5)); [/asy][/asy] Thus, $W\in\overline{W_1W_2}$. Similarly, midpoint $X,Y,Z$ of $BC,CD,DA$, respectively lie on a side of $\mathcal{P}$. Now, we consider another parallelogram $\mathcal{R}'$ which is symmetric across $O$, and whose each side subtend an angle of $45^{\circ}$ to the sides of $\mathcal{R}$, and with minimal area. Clearly, the sides of $\mathcal{R'}$ must pass through a vertex. Let $A_1, B_1, C_1, D_1$ be those vertices in the northwest, northeast, southeast, and southwest, respectively. [asy][asy] size(6cm); defaultpen(fontsize(10pt)); pair A = (0,1); pair B = (1,1); pair C = (1,0); pair D = (0,0); pair W = (A+B)/2; pair X = (B+C)/2; pair Y = (C+D)/2; pair Z = (D+A)/2; pair B_1 = (0.85,0.85); pair A_1 = (0.25,0.85); pair C_1 = (1,1)-A_1; pair D_1 = (1,1)-B_1; pair W_1 = extension(A_1,A_1+(1,1),B_1,B_1+(1,-1)); pair X_1 = extension(C_1,C_1+(1,1),B_1,B_1+(1,-1)); pair Y_1 = (1,1)-W_1; pair Z_1 = (1,1)-X_1; filldraw(W--B_1--X--C_1--Y--D_1--Z--A_1--cycle,yellow,dotted); draw(A--B--C--D--cycle,linewidth(1.5)); draw(W_1--X_1--Y_1--Z_1--cycle); dot('$A$',A,NW); dot('$B$',B,NE); dot('$C$',C,SE); dot('$D$',D,SW); dot('$W$',W,dir(75)); dot('$X$',X,dir(10)); dot('$Y$',Y,S); dot('$Z$',Z,dir(180)); dot('$A_1$',A_1,NW); dot('$B_1$',B_1,NE); dot('$C_1$',C_1,SE); dot('$D_1$',D_1,SW); label("\Large$\mathcal{P}$",(0.5,0.5)); [/asy][/asy] Let $AB=2$, and let the length of the sides of $\mathcal{R}'$ that contains $A_1,B_1$ are $2a, 2b$ respectively; therefore $ab\geq 1$ by minimality. Then, we can bound the area of $\mathcal{P}$ as \begin{align*} [\mathcal{P}] &\geq [WXYZ] + 2([A_1WZ] + [B_1WX]) \\ &= 2 + 2\left(\frac 12 \cdot \sqrt{2}\cdot \left(b-\frac{\sqrt 2}{2}\right) + \frac 12 \cdot \sqrt{2}\cdot \left(a-\frac{\sqrt 2}{2}\right)\right) \\ &= \sqrt{2}(a+b) \\ &\geq 2\sqrt{2} \end{align*}or $ABCD$ already satisfies the condition.

We begin by generalizing the problem to smooth closed convex curves $\partial\mathcal{P}$ symmetric about $O$ with interior $\mathcal{P}$. For any point $A\in\partial\mathcal{P}$, let $t(A)$ be the line tangent to $\partial\mathcal{P}$ at $A$, and let $-A$ be the symmetric point about $O$. Perturb $\mathcal{P}$ by an infintesmal amount so that $\partial\mathcal{P}$ has no inflection points, or that the rate of change of the direction of $t(A)$ is never $0$. Note that $t(A)\parallel t(-A)$. For any points $A,B\in\partial\mathcal{P}$, let $\mathcal{R}(A,B)$ be the parallelogram bounded by the lines $t(A),t(B),t(-A),t(-B)$. Note that \[\mathcal{R}(A,B) = \mathcal{R}(\pm A,\pm B).\]We suppose that the problem is false, so we have \[[\mathcal{R}(A,B)]> \sqrt{2}\cdot [\mathcal{P}]\]for all $A,B\in\partial\mathcal{P}$. Note that $(A,B)\mapsto [\mathcal{R}(A,B)]$ is a continuous function from the compact set $\partial\mathcal{P}\times\partial\mathcal{P}$ to $\mathbb{R}$, so there exists some pair $(A_m,B_m)$ for which $[\mathcal{R}(A,B)]$ is minimized. We have the following crucial claim. Claim: We have $OA_m\parallel t(B_m)$ and $OB_m\parallel t(A_m)$. Proof: To simplify the calculation, we first show that there exists some $X\in\partial\mathcal{P}$ such that $OX\perp t(X)$. Indeed, by compactness, we may pick $X$ such that $d(X,-X)$ is maximized, and it is easy to see that if $OX$ is not perpendicular to $t(X)$, then there is some direction we can slide $X$ along $\partial\mathcal{P}$ to increase $d(X,-X)$. Now, let $OX$ be the $x$-axis, with $O$ the origin. Let $(x,f(x))$ be the curve $\partial\mathcal{P}$ in the upper half plane. We know $f$ exists since as $A$ varies from $-X$ to $X$ in the upper half plane, the slope of $t(A)$ varies monotonically from $-\infty$ to $\infty$, so we can integrate to find $f$, and there won't be any issues since $\infty$ is never hit besides at the boundary. By potentially replacing $A_m$ with $-A_m$ and $B_m$ with $-B_m$, we may assume that $A_m=(x_1,f(x_1))$ and $B_m=(x_2,f(x_2))$ for some $x_1,x_2$. We first assume that $f(x_1),f(x_2)\ne 0$ (i.e. $A_m,B_m\not\in\{X,-X\}$).

Figure

Let $(A,B)$ vary in some small open neighborhood of $(A_m,B_m)$, and abuse notation by letting $A=(x_1,f(x_1))$ and $B=(x_2,f(x_2))$. Let $P$ be the point on $t(B)$ such that $OP\parallel t(A)$, and let $Q$ be the point on $t(A)$ such that $OQ\parallel t(B)$. Finally, let $T=t(A)\cap t(B)$. Note that \[[\mathcal{R}(A,B)] = 4\cdot[OPTQ].\]It is not hard to check that \[P=\left(\frac{x_2f'(x_2)-f(x_2)}{f'(x_2)-f'(x_1)},\,\,\,f'(x_1)\cdot\frac{x_2f'(x_2)-f(x_2)}{f'(x_2)-f'(x_1)}\right)\]and \[Q=\left(\frac{x_1f'(x_1)-f(x_1)}{f'(x_1)-f'(x_2)},\,\,\,f'(x_2)\cdot\frac{x_1f'(x_1)-f(x_1)}{f'(x_1)-f'(x_2)}\right),\]and thus that \[[OPTQ]=\left|\frac{[x_1f'(x_1)-f(x_1)]\cdot[x_2f'(x_2)-f(x_2)]}{[f'(x_1)-f'(x_2)]^2}\right|.\]Clearly the value of $[OPTQ]$ is continuous, and nonzero at $(A_m,B_m)$, so we may select our neighborhood small enough such that the absolute value imparts a consistent sign over the neighborhood. Since $(A_m,B_m)$ is a local maximum, we must have that \[\frac{\partial}{\partial x_1}[OPTQ] = \frac{\partial}{\partial x_2}[OPTQ] = 0.\]Supposing WLOG that the sign imparted by the absolute value is positive, we see that \begin{align*} \frac{\partial}{\partial x_1}[OPTQ] &= [x_2f'(x_2)-f(x_2)]\cdot\frac{x_1f''(x_1)[f'(x_2)-f'(x_1)] - [x_1f'(x_1)-f(x_1)]\cdot(-f''(x_1))}{[f'(x_1)-f'(x_2)]^2} \\ &= [x_2f'(x_2)-f(x_2)]\cdot\frac{f''(x_1)\cdot[x_1f'(x_2)-f(x_1)]}{[f'(x_1)-f'(x_2)]^2}. \end{align*}Since $[OPTQ]$ is nonzero, we see that $x_2f'(x_2)-f(x_2)$ is nonzero, and furthermore $f''(x_1)$ is nonzero since we assumed that $\partial\mathcal{P}$ has no inflection points. Therefore, we have \[f'(x_2)=f(x_1)/x_1,\]which says exactly that $OA_m\parallel t(B_m)$. We may derive the other relation in a similar manner. If say $A_m=-X$, we may do a similar proof, where we now only vary $B$ in a neighborhood of $B_m$, and a similar calculation yields that $t(B_m)\parallel OX$, and thus $OB_m\perp OX$ by Rolle's theorem. This completes the proof of the claim. $\blacksquare$ Let $T=t(A_m)\cap t(B_m)$. The above claim implies that $OA_mTB_m$ is a parallelogram, and that \[[\mathcal{R}(A_m,B_m)] = 4\cdot[OA_mTB_m].\]Take an affine transformation so that $OA_mTB_m$ is a unit square, and let $\{Y,-Y\}=\ell\cap\partial\mathcal{P}$, where $\ell$ is the line through $O$ parallel to $A_mB_m$. Note that $A_m$ and $B_m$ are on the same arc bounded by $Y$ and $-Y$, so WLOG that they are both above $OY$. Let $H$ be the unique point above $OY$ on $\partial\mathcal{P}$ such that $t(H)\parallel OY$. Furthermore, note that $\partial\mathcal{P}$ is tangent to $A_mT$ and $B_mT$ at $A_m$ and $B_m$, so $H$ is above $A_mB_m$.

Figure

We see that $\mathcal{P}$ is convex, so its upper half contains the convex hull of $\{-Y,A_m,H,B_m,Y\}$, which has area \[\frac{1}{2}y\frac{1}{\sqrt{2}}\cdot 2+\frac{1}{2} + \frac{1}{2}\sqrt{2}\left(h-\frac{1}{\sqrt{2}}\right) = \frac{y+h}{\sqrt{2}},\]so \[[\mathcal{P}]\ge \sqrt{2}\cdot(y+h).\]We have \[4=[\mathcal{R}(A_m,B_m)]> \sqrt{2}\cdot [\mathcal{P}] = 2(y+h),\]so $y+h<2$. Now, \[[\mathcal{R}(H,Y)]\le 4yh,\]so we also have \[4yh>2(y+h),\]or $y+h<2yh$. Thus, by AM-HM \[\frac{y+h}{2}\ge \frac{2yh}{y+h}>1,\]which contradicts $y+h<2$. This is the desired contradiction, so the proof is complete.