This question related to maxima and minima concept so we have to use first and second derivative test.
first we have to asume that number so,
let x is required number
according the condition x + 1/x=as small as possible
now we make one function of x and it's reciprocal
take first derivative of f(x)
put f '(x)=0 to get maximum and minimum value
so one is maximum value and one is minimum value, to find which one is minimum we have to test now second derivative test, so
put x=1 in f ''(x)
f ''(1)=2>0 (greater than zero mean it's minimum)
f ''(-1)=-2<0 (less than zero mean it's maximum)
so x=1 is required answer .
according this you can solve easily second part.