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.

Remark
$$\displaystyle\sum f(a,b,c)=f(a,b,c)+f(b,c,a)+f(c,a,b)$$
means cyclic sum.
$$\displaystyle\prod f(a,b,c)=f(a,b,c)\cdot f(b,c,a) \cdot f(c,a,b)$$
means cyclic product.

$$
\left( \frac{b+c}{a}+2 \right) ^2+\left( \frac{c}{b}+2 \right) ^2+\left( \frac{c}{a+b}-1 \right) ^2
$$
$$
=\left( \frac{a+b+c}{a}+1 \right) ^2+\left( \frac{b+c}{b}+1 \right) ^2+\left( \frac{c}{a+b}-1 \right) ^2
$$
then we can make the substitution $\left( a+b+c,b+c,c \right) \mapsto \left( x,y,z \right) $ to get a pretty form
$$
=\left( \frac{x}{x-y}+1 \right) ^2+\left( \frac{y}{y-z}+1 \right) ^2+\left( \frac{z}{z-x}+1 \right) ^2
$$
then make a substitution again
$$
\left( \frac{x}{x-y},\frac{y}{y-z},\frac{z}{z-x} \right) \mapsto \left( a,b,c \right) 
$$
and try to find an equality
$$
a=\frac{x}{x-y}\Rightarrow \frac{a-1}{1}=\frac{y}{x-y}\Rightarrow \frac{a-1}{\left( a-1 \right) +1}=\frac{y}{\left( x-y \right) +y}\Rightarrow \frac{a-1}{a}=\frac{y}{x}
$$
$$
\prod{\frac{a-1}{a}}=1\Rightarrow \prod{\left( a-1 \right)}=abc\Rightarrow \sum{a}=\sum{ab}+1
$$
then we can get final conclusion
$$
\sum{\left( a+1 \right) ^2}=\sum{a^2}+2\sum{a}+3=\sum{a^2}+2\sum{ab}+6=\left( \sum{a} \right) ^2+5\geq 0
$$
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there is a similar problem with similar tricks
Assuming $\displaystyle \sum x=0$. Find the minimum of $\displaystyle \sum \left(\frac{x}{y}\right)^2$. (Answer : 5)
===============Solution=============
$$
\begin{aligned}
&\sum{\left( \frac{x}{y} \right) ^2}\\
=&\left( \sum{\frac{x}{y}} \right) ^2-2\sum{\left( \frac{x}{y}\cdot \frac{y}{z} \right)}\\
=&\left( \sum{\frac{x}{y}} \right) ^2+2\sum{\frac{-x}{z}}\\
=&\left( \sum{\frac{x}{y}} \right) ^2+2\sum{\frac{y+z}{z}}=\left( \sum{\frac{x}{y}} \right) ^2+2\sum{\frac{x}{y}}+6\\
=&\left( \sum{\frac{x}{y}}+1 \right) ^2+5\\
\geq & 5
\end{aligned}
$$
Of course you can make it into perfect square.
$$\frac{x^2}{y^2}+\left( \frac{x+y}{x} \right) ^2+\left( \frac{y}{x+y} \right) ^2=\frac{\left( x^3+x^2y-2xy^2-y^3 \right) ^2}{x^2y^2\left( x+y \right) ^2}+5\geq 5
$$
