How prove $\left(\frac{b+c}{a}+2\right)^2+\left(\frac{c}{b}+2\right)^2+\left(\frac{c}{a+b}-1\right)^2\ge 5$ Let $a,b,c\in R$ and $ab\neq 0,a+b\neq 0$. Find the minimum of:
$$\left(\dfrac{b+c}{a}+2\right)^2+\left(\dfrac{c}{b}+2\right)^2+\left(\dfrac{c}{a+b}-1\right)^2\ge 5$$
if and only if $$a=b=1,c=-2$$
My idea: Since 
$$\left(\dfrac{b+c+2a}{a}\right)^2+\left(\dfrac{c+2b}{b}\right)^2+\left(\dfrac{c-a-b}{a+b}\right)^2$$
let
$$x=\dfrac{b+c+2a}{a},y=\dfrac{c+2b}{b},z=\dfrac{c-a-b}{a+b}$$
then I can't work.
Thank you.

Continuing from Calvin Lin' answer, the minimum value will be otained when all partial derivatives will be zero. The only partial derivative which is easy to write is with respect to $c$ which only appears in numerator. Almost as Calvin Lin wrote, this derivative cancels when $$c=-\frac{b (a+b) (2 a+b)}{a^2+a b+b^2}$$ Back to the expression, this leads to a constant value of 5.
The expansion is not so tedious since, replacing $c$ by its expression, we can find $$\frac{b+c}{a}+2=-\frac{(a+b) (a+2 b)}{a^2+a b+b^2}$$ $$\frac{c}{b}+2=\frac{b (b-a)}{a^2+a b+b^2}$$  $$\frac{c}{a+b}-1=\frac{a (2 a+b)}{a^2+a b+b^2}$$
