Prove that at a wedding reception you don't need more than $20 \sqrt{mn}$ of ribbon to adornate the cakes. 
At a wedding reception,$n$ guests have assembled into $m$ groups to
  converse.(The groups are not necessarily equal sized.)The host is
  preparing $m$ square cakes,each with an ornate ribbon adorning its
  perimeter,to serve the $m$ groups.No guest is allowed to have more
  than $25 $ $cm^2$ of cake.Prove that no more than $20 \sqrt{mn} $ $cm$
  of ribbon is needed to embellish the $m$ cakes.

I am quite clueless on how this kind of problem has to be tackled.However below are my  thoughts:
I define $x_1+x_2+\cdots +x_m$ to be the number of people in group $i$ ,$1 \le i \le m$, and $s_1+s_2+\cdots+s_m$  to be the side length of the the squared cake in the rispective groups.
Now I should prove that $4(s_1+s_2+\cdots+s_m)\le 20 \sqrt{mn}$ so the problem I face now is to expresseach of $s_i$ in terms of the numbers of guests in the $m$ groups. I don't see how to do that.

You are almost there
$$s_i^2=25 x_i \implies \sum_{i=1}^ms_i^2=25 n$$
$$L= 4 \sum_{i=1}^m s_i$$
$$20 \sqrt{mn}=20 \sqrt{m \frac{\sum_{i=1}^m s_i^2}{25}}= 4 \sqrt{m \sum_{i=1}^m  s_i^2} $$
Hence we want to prove that 
$$ \sum_{i=1}^m s_i \le \sqrt{m \sum_{i=1}^m  s_i^2} $$ 
or 
$$ \left(\frac{\sum_{i=1}^m s_i}{m}\right) ^2 \le \frac{1}{m} \sum_{i=1}^m  s_i^2 $$ 
which is true (statistical version of the well known probabilistic inequation $(E[X])^2\le E[X^2]$), can be proved by AM-GM or by Jensen inequality.

Update:
Let $\bar{y}=\sum y_i/m$ 
Then $0\le \sum (y_i -\bar{y})^2 =\sum y_i^2 - 2 \bar{y}\sum y_i + m\bar{y}^2=\sum y_i^2 -  m\bar{y}^2 $. Hence
$$ \frac{\sum y_i^2}{m} \ge  \bar{y}^2  = \left( \frac{\sum y_i}{m}\right)^2$$
