If $A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}\;,$ Then $\lfloor A \rfloor\;\;,$ 
If $$A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}\;,$$ Then $\lfloor A \rfloor\;\;,$ Where $\lfloor x \rfloor $ Represent floor fiunction of $x$

My Try:: Using $\bf{A.M\geq H.M\;,}$ We get
$$\frac{1980+1891+1982+....+2012}{33}>\frac{33}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}$$
So $$\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}<\frac{1980+1981+....+2012}{(33)^2}=\frac{1996}{33}\approx 60.5<61$$
Now how can i prove that the above expression $A$ is $>60$
Help me, Thanks 

Let $$A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}=\frac 1K$$
$$K=\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}<\frac{1}{1980}+\frac{1}{1980}+\frac{1}{1980}+........+\frac{1}{1980}=\frac{33}{1980}=\frac{1}{60}$$
$$A=\frac 1K >60$$
