Fall 2018, problem 70
Prove that for any convex quadrilateral of area 1, the sum of the lengths of all sides and diagonals is not less than $4+\sqrt{8}$.
Comments
The claim will obviously follow from the following points 1 and 2:

For any quadrilateral of area 1 there exists a rhombus of area 1 for which the sum is less.

For any rhombus of area 1 the sum is greater than the one of unit square.
Property 1 has a simple geometrical proof.
Let ABCD be a quadrilateral of area 1; choose a diagonal (e.g. AC) and let $r_1,r_2$ be the lines parallel to AC passing throug B, D respectively; we call a,b the sides on the halfplane of B and c,d the other two sides.
Moving B along $r_1$ and D along $r_2$ the area remains the same; the length of BD is minimal when BD is orthogonal to AC and both the sums a+b, c+d are minimal when BD cuts AC on its middle point; thus with a=b, c=d.
Do the same with respect to the diagonal BD we get a=d, b=c.
Unfortunateli, concerning property 2, I do not see a simple geometrical proof; however the algebric proof is a (tedious but) trivial inequality.
 S=a+b+c+d. Â Â T=a+b+c+d+m+n
 let a,b,c,d be sides of a quadrilateral and m and n are lengths of diagonals.
 The angle between diagonals be â‚¬ and between sides of lengths a and d be Î± and between b and c be Î².(a,b,c,d are consecutive).
 1=m nsin(â‚¬)/2...m n=2cosec(â‚¬)...mn>=2.
 1=[a dsin(Î±)+b csin(Î²)]/2
 since a d>=ad sin(Î±) and bc>=b csin(Î²)...
 a d+bc >=2
 let a d=x^2 and bc=y^2. We have to find min. of a+b+c+d+m+n since n>=2/m (adding m both the sides) m+n>=m+2/m...m+n >=2âˆš2(AM GM enequality)and write a+b+c+d as (a+d)+(b+c). Again a+d>=2(âˆša âˆšd) Â Â Â Â Â b+c>=2(âˆšbâˆšc) ...adding these S>=2(x+y). We have to minimize x+y given x^2+y^2>=2 Off course it will be then when line x+y=c(say) will tangent to circle x^2+y^2=2(note that x and y are positive ) it gives c=2 . hence S>=4 and m+n >=âˆš8 Which will give T>=4+âˆš8
Coolidge's formula for the area of a convex planar quadrilateral is:
$K^2 = \frac{(a^2 + b^2 + c^2 + d^2)^2 + 8abcd 2(a^4 + b^4 + c^4 +d^4) 4(ac + bd +pq)(ac +bd pq)}{16}$
where $K$ is the area and $a, b, c, d, p, q$ are the lengths of the 4 sides and the two diagonals, respectively.
The problem is to find the minimum of the function:
$a + b + c + d + p + q$
subject to the constraint that the above area equals 1. (Obviously, geometric lengths are nonnegative, so that, additionally, all these variables are greater than or equal to zero.) So, just use the Method of Langrange Multipliers for constrained optimization problems from Advanced Calculus. Define:
$h(a,b,c,d,p,q) = a + b + c + d + p + q  m( (a^2 + b^2 + c^2 + d^2)^2 + 8abcd 2(a^4 + b^4 + c^4 +d^4) 4(ac + bd +pq)(ac +bd pq)16$)
This function is to be minimized over the the nonnegative values of the six variables. (Help me out  what generalizes "octant" in this context?) Anyway, just take the six partial derivatives and set them to zero.
Clearly, these equations are symmetric in $a, b, c, d$ and also in $p, q$. Common sense suggests taking the
$p, q$ derivatives first, and, eliminating $m$ (as usual), quickly gives $p = q$. After taking two of the $a, b, c, d$
partial derivatives (really, just one with the symmetry) and eliminating $m$ gives an algebraic equation quite
close to having, say, $(a  b)$ as a factor, where the only problem is another term with $(c  d)$ as a factor.
This should suggest $a = b$ and $c = d$, and if you carry on solving in this way, you can show that, and
eventually $a = b = c = d$. Consequently, the minimizing quadrilateral is a square and the result follows
immediately.
Let $ABCD$ be a convex quadrilateral of unit area. By the Isoperimetric Theorem, among all quadrilaterals of a given area $\Rightarrow$ the square has the smallest perimeter. This implies $AB = BC = CD = DA = 1$, and the perimeter:
$AB + BC + CD + DA \geq 4$ (i).
Also, the area of $ABCD$ equals $\frac{1}{2} \cdot AC \cdot BD \cdot \sin(\theta)$, where $\theta$ is the included angle between diagonals $AC$ and $BD$. This is exactly half the area of the parallelogram $ACC'A'$, which is formed on the original diagonals $AC$, $BD$. Thus, the area of $ACC'A'$ equals $2$, which is a square of side length $\sqrt{2}$ and a minimum perimeter of $4\sqrt{2}$. This in turn yields:
$2(AC+BD) \geq 4\sqrt{2} \Rightarrow AC + BD \geq 2\sqrt{2}$ (ii).
Combining inequalities (i) & (ii) produces:
$(AB + BC + CD + DA) + (AC + BD) \geq 4 + 2\sqrt{2}$
with equality occuring when $ABCD$ is a square.
$\mathbb{Q}. \mathbb{E}. \mathbb{D}.$