Eukleidis wrote: Let ABCD be a convex quadrilateral and points E and F on sides AB,CD such that \[\tfrac{AB}{AE}=\tfrac{CD}{CF}=n\] If S is the area of AEFD show that ${S\leq\frac{AB\cdot CD+n(n-1)AD^2+n^2DA\cdot BC}{2n^2}}$ Let it be rectangle with $AB=CD=100,AD=BC=1$ and $n=100$ than we have $50\le 0.5+\frac{2n^2-n}{2n^2}<1.5$ hence a mistake in question?

I made a mistake. Its not CF but DF. Sorry.

There should be ${nDA\cdot BC}$ instead of ${n^2DA\cdot BC}$ it seems. We can get it using Ptolemy's inequality ${AF\cdot ED\leq AE\cdot DF+AD\cdot EF}$ and other noname inequality $\sqrt{(a+b)^2 +(c+d)^2}\leq\sqrt{a^2 +c^2 }+\sqrt{b^2 +d^2 }.$

Indeed, it is just an exercise in known formulae (by the way, that "noname" inequality is Minkowski's, also known as the "triangle" inequality ). And indeed, the $n^2DA\cdot BC$ should be $nDA\cdot BC$; mind you, when it works for the latter, it will a fortiori work for the former! The area $S$ of the quadrilateral $AEFD$ is $S = \dfrac {1} {2} AF\cdot DE\cdot \sin\angle(AF,DE) \leq \dfrac {1} {2} AF\cdot DE \leq \dfrac {1} {2} (AE\cdot DF + AD\cdot EF)$, by the known fact that the area of a convex quadrilateral is half the product of its diagonals with the sine of their angle, followed by the fact that the sine of an angle is at most $1$, followed by Ptolemy's inequality. Now, $AE\cdot DF = \dfrac {AB\cdot DC} {n^2}$, so if we prove that $AD\cdot EF \leq \dfrac {(n-1)AD^2 + AD\cdot BC} {n}$, i.e. $EF \leq \dfrac {(n-1)AD + BC} {n}$, that will be enough. But this is a simple vectorial computation. Denote by lowercase letters the position vectors of the uppercase points. Then $e = a+\dfrac {1} {n} (b-a)$ and $f = d+\dfrac {1} {n} (c-d)$, so $e-f = (a-d) + \dfrac {1} {n} ((b-c) - (a-d)) = \dfrac {n-1} {n} (a-d) + \dfrac {1} {n} (b-c)$. Then $EF = |e-f| =$ $ \left |\dfrac {n-1} {n} (a-d) + \dfrac {1} {n} (b-c)\right | \leq$ $ \dfrac {n-1} {n} |a-d| + \dfrac {1} {n} |b-c| =$ $ \dfrac {n-1} {n} AD + \dfrac {1} {n} BC$, by the triangle inequality.

Quote: also known as the "triangle" inequality I know but "triangle" isn't a name actually. Oh well, just kidding

iron wrote: and other noname inequality $\sqrt{(a+b)^2 +(c+d)^2}\leq\sqrt{a^2 +c^2 }+\sqrt{b^2 +d^2 }.$ mavropnevma wrote: Indeed, it is just an exercise in known formulae (by the way, that "noname" inequality is Minkowski's, also known as the "triangle" inequality It's not quite Minkowski in the standard form in which that is usually quoted, and why is it the triangle inequality? It is, however, straight AM/GM if you just square each side a couple of times. Merlin

Consider points $O(0,0), A(a,c), B(-b,-d)$. Then $\sqrt{(a+b)^2 + (c+d)^2} = AB$, $\sqrt{a^2 + c^2} = AO$, $\sqrt{b^2 + d^2} = OB$, so the inequality is equivalent to $AB \leq AO + OB$, which, as far as I remember, is the triangle inequality, expressed in the Euclidean norm (in any dimension) by Minkovski's inequality. The fact it is not appearing in its usual form $\sqrt{(x_1-x_3)^2 + (y_1-y_3)^2} \leq$ $ \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2} + \sqrt{(x_2-x_3)^2 + (y_2-y_3)^2}$, but is just a particular form, does not rob it of its origin.

Marius Stanean's proof Take points $X,Y$ such as $AX\parallel CD \wedge FX\parallel DA,\; BY\parallel CD \wedge FY\parallel BC\Rightarrow ADFX,BCFY$ are parallelograms. We have $\left.\begin{array}{l} \frac{AX}{BY}=\frac{DF}{FC}=\frac{1}{n-1}=\frac{AE}{BE}\\ AX\parallel BY\end{array}\right\}\Rightarrow \triangle AXE\sim\triangle BYE\Rightarrow X,E,Y$ collinear $\frac{EX}{EY}=\frac{1}{n-1}\,.$ Fie $Z\in FE,\;(E\in(FZ))$ such as $\frac{EF}{EZ}=\frac{1}{n-1}\Rightarrow \triangle EFX\sim\triangle EZY\Rightarrow \frac{FX}{ZY}=\frac{1}{n-1}$. In triangle $ZFY$ $FZ\le FY+ZY\Leftrightarrow \boxed{n\cdot EF\le BC+(n-1)AD}\;\;\;(*)$ $S=\frac{AF\cdot DE}{2}\cdot \sin\angle (AF,DE)\le \frac{AF\cdot DE}{2}\stackrel{\mbox{\tiny I.Ptolemeu}}{\le} \frac{AD\cdot EF+AE\cdot DF}{2}=$ $\frac{n^2\cdot AD\cdot EF+AB\cdot CD}{2n^2}\stackrel{(*)}{\le}\frac{AB\cdot CD+n(n-1)DA^{2}+n\cdot DA\cdot BC}{2n^2}\,.$

So this proof offers a synthetic argument for $EF \leq \dfrac {(n-1)AD + BC} {n}$ (having as last recourse the triangle inequality - what else?); otherwise it is difficult to foresee a change in the chain of formulae.

The solution is originaly from http://forum.gil.ro/viewtopic.php?f=19&t=1198 .I don't understand what part of the proof is incomplete?

No part is incomplete (never said that) - it's just that the chain of ideas must be the same.

Eukleidis wrote: Let ABCD be a convex quadrilateral and points E and F on sides AB,CD such that \[\tfrac{AB}{AE}=\tfrac{CD}{DF}=n\] If S is the area of AEFD show that ${S\leq\frac{AB\cdot CD+n(n-1)AD^2+nDA\cdot BC}{2n^2}}$