Looks tough but easy

Algebra Level 3

Find $a+b+c$ when $a = 3$ in the following equation:

$a^{2}+b^{2}+c^{2} = (3 \sqrt[3]{abc})^{2} -2(ab+bc+ac)$

Note: $a,b,c$ are positive integers.

1 solution

Mohammad Al Ali
May 24, 2014

The given equation can be rewritten as, $a^{2}+b^{2}+c^{2}+2ab+2bc+2ac= (3 \sqrt[3]{abc})^{2}$ .

The LHS can be simplified into $(a+b+c)^{2}$ .

Hence the equation now looks like, $(a+b+c)^{2} = (3 \sqrt[3]{abc})^{2}$

Taking square root from both sides we obtain,

$a+b+c = 3 \sqrt[3]{abc}$

Or simply, $\frac{a+b+c}{3} = \sqrt[3]{abc}$

According to the Arithmetic Mean-Geometric Mean Inequality , this equality only holds if and only if all members of the set are equal.

Hence we can conclude that $a = b = c = 3$ , thus $\boxed{a+b+c = 9}$ .

Note that in order to apply AM-GM, you have to ensure that the values are non-negative.

Calvin Lin Staff - 5 years, 11 months ago