Question

Given $a+b+c=2$, $a,b,c \in R^+$. Find minimum value of $\frac{1}{a} + \frac{4}{b} + \frac{36}{c}$.

Answer 1

$\frac{81}{2}$

Answer 2

Apply Cauchy-Schwarz inequality in Engel form: $\sum \frac{y_i^2}{x_i} \ge \frac{(\sum y_i)^2}{\sum x_i}$. With $y_1=1, y_2=2, y_3=6$ and $x_1=a, x_2=b, x_3=c$. The expression is $\frac{1^2}{a} + \frac{2^2}{b} + \frac{6^2}{c} \ge \frac{(1+2+6)^2}{a+b+c}$. Using the constraint $a+b+c=2$, the minimum value is $\frac{9^2}{2} = \frac{81}{2}$.