Answer to Question #247362 in Algebra for M.l

Question

What is the smallest value of 9(a^2) + (b^2) +16(c^2) given that 3a-b+4c=16 and

(4/a) - (12/b) +(3/c) = 0

Answer 1

Given that: $3a-b+4c=16\ldots(1)$

Also, $\frac{4}{a} - \frac{12}{b} +\frac{3}{c} = 0$

$\Rightarrow \frac{4bc-12ac+3ab}{abc}=0\\ \Rightarrow 4bc-12ac+3ab=0\ldots(2)$

Squaring equation $(1),$ we get:

$9a^2+b^2+16c^2+2(-3ab-4bc+12ac)=0\\ \Rightarrow 9a^2+b^2+16c^2-2(3ab+4bc-12ac)=0\\ \Rightarrow 9a^2+b^2+16c^2=0 \ [From \ (2)]$