Bound on number of embassies given that no three countries have pairwise relations 
There are 20 countries on the planet,among any three of these countries there are always two with no diplomatic relations. Prove that there are at most 200 embassies on the planet.

I tried using the extremal principle, getting country with the biggest number of diplomatic relations but had no success in proving the problem.
 A: To make it clear, I will adjust the proof of Mantel's theorem to this problem. (Credit goes to Ross Millikan for the idea).

There are $m$ countries, so that for any three of them there are at least two having no diplomatic relations with each other.
$d$ - total number of diplomatic relations
Prove that $d \le \frac{m^2}{4}$

Take a country $A$ with maximum number of embassies, denoted by $n$ (i.e. none of others has more than $n$ embassies). 
$N$ is the set of $n$ countries having embassies in $A$.
$M$ is the complement set (i.e. countries not contained in $N$) of $m-n$ countries.
Neither of two $N_1$ and $N_2$ from $N$ may have diprelations with each other, otherwise any of $A$, $N_1$ and $N_2$ would have diprelations with each other, which is disallowed.
Therefore a diprelation may be one of two types:
"Internal": between two $M$-countries, total number $d_1$
"External": between an $M$-country and an $N$-country, total number $d_2$. 
$d_1 + d_2 = d$ (total number of diprelations)
Estimate total number of embassies in $M$-countries. Each internal relation results in $2$ embassies in $M$, while each external relation gives $1$ embassy in $M$ (another in $N$). Therefore   
$E_M = 2d_1 + d_2 \ge d_1 + d_2 = d \quad (1)$
On the other side, due to the choice of country $A$, each $M$-country has no more than $n$ embassies. Hence:
$E_M \le (m-n)n \le [\frac{(m-n)+n}{2}]^2 = \frac{m^2}{4} \quad (2)$
Comparing $(1)$ and $(2)$ gives: $d \le \frac{m^2}{4}$, q.e.d.

In particular if $M=20$, then $d\le100$, so no more than $200$ embassies.

PS. Show that this estimation cannot be improved. 
If $m$ is even, split countries into two sets $M$ and $N$, $\frac{m}{2}$  countries in each, and let them have only "external" relations (between $M$-country and $N$-country, but not within $M$ or $N$). So there are $\frac{m^2}{4}$ diprelations, and each three countries have a non-related pair.
If $m$ is odd, then sets $M$ and $N$ will have $\frac{m-1}{2}$ and $\frac{m+1}{2}$  countries, so there will be $\frac{m^2-1}{4}$ diprelations, i.e. the integral part of $\frac{m^2}{4}$.
