Let #a=b=c=1/3#. Then the expression evaluates to #100/3#, the lower limit of the inequality.
Perturb one value by a positive amount #p#. Then another value sees the opposite change #-p#. Then the sum is
#(a+1/a)^2+(a+p+1/(a+p))^2+(a-p+1/(a-p))^2#
#=(a+1/a)^2+(a+p)^2+2+1/(a+p)^2+(a-p)^2+2+1/(a-p)^2#
#=(a+1/a)^2+a^2+2ap+p^2+2+1/(a+p)^2+a^2-2ap+p^2+2+1/(a-p)^2#
#=(a+1/a)^2+2(a^2+2+p^2)+((a-p)^2+(a+p)^2)/((a-p)^2(a+p)^2)#
#=(a+1/a)^2+2(a^2+2+p^2)+2(a^2+p^2)/((a^2-p^2)^2)#
Multiplying out an original bracket gives us:
#a^2+2+1/a^2#
The second term of our perturbed sum is greater by #2p^2# than the sum of the portions of the second and third original terms that don't include the #1/a^2# term.
The third term of our perturbed sum is greater than the sum of the two #1/a^2# terms from the second and third original terms, because, for positive #p^2 < a^2#, #2/a^2=2a^2/(a^2)^2<2(a^2+p^2)/(a^2)^2<2(a^2+p^2)/(a^2-p^2)^2#.
Thus #(a+1/a)^2+(a+p+1/(a+p))^2+(a-p+1/(a-p))^2>3(a+1/a)^2# and we have proven that #a=b=c=1/3# are the value choices that make the expression take its least value of #100/3# - any perturbation of the variables from this results in an increase in the sum.
