<s>[INST] 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.
 [/INST] Note that "if and only if" is clearly not true, since any multiples will work as your expression is homogenous. In fact, the equality case is more general than what you listed.
Differentiating with respect to $c$ (which only appears in the numerator, hence is not too ugly), you can show that (Thanks Wolfram) the minimum occurs when
$$ c = - \frac{ b^3+3ab^2+2a^2b } {a^2+ab+b^2 }. $$
Substituting this back into the expression (and tediously expanding), we get the value of 5. Hence, the minimum is 5. 
Now, verify that this is indeed the global minimum by doing all the proper checks (I'm too lazy to complete this calculus approach completely.)
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