what is the smallest value of $a+b+c$

Question

Set a, b, and c to be positive numbers, and $\frac{4}{a} + \frac{9}{b} + \frac{4}{c} =1$. Find the smallest value of $a+b+c$.

I think this problem is all about AM-GM, but i dont see any strategies to apply AM-GM. Any help? Thanks in advance

Answer 1

By Cauchy-Schwarz Inequality, we have: $\displaystyle (\frac{4}{a}+\frac{9}{b}+\frac{4}{c})(a+b+c)\ge(2+3+2)^2=49$. Hence, $a+b+c\ge49$.

To add as an proof to the Cauchy-Steward inequality for 3 terms, we have
$(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)\ge(a_1b_1+a_2b_2+a_3b_3)^2$

Proof. Expanding the inequality yields:
$a_1^2b_2^2-2a_1a_2b_1b_2+a_2^2b_1^2+a_1^2b_3^2-2a_1a_3b_1b_3+a_3^2b_1^2+a_2^2b_3^2-2a_2a_3b_2b_3+a_3^2b_2^2\ge0$
$\Rightarrow(a_1b_2-a_2b_1)^2+(a_1b_3-a_3b_1)^2+(a_2b_3-a_3b_2)^2\ge0$
which is trivial.