Let a,b,c be positive real numbers such that a+b+c=1.Prove the following inequality: (a+1/a)^2+(b+1/b)^2+(c+1/c)^2>=100/3?

1 Answer
Jun 14, 2018

Set all parameters equal to achieve the inequality limit. Then demonstrate that perturbing one of them results in an increase in the sum.

Explanation:

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.