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.
 A: 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.)
A: 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
$$
\sum{\left( a+1 \right) ^2}=\sum{a^2}+2\sum{a}+3=\sum{a^2}+2\sum{ab}+6=\left( \sum{a} \right) ^2+5\geq 0
$$
==================================
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.
$$\frac{x^2}{y^2}+\left( \frac{x+y}{x} \right) ^2+\left( \frac{y}{x+y} \right) ^2=\frac{\left( x^3+x^2y-2xy^2-y^3 \right) ^2}{x^2y^2\left( x+y \right) ^2}+5\geq 5
$$
A: 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 $$\frac{b+c}{a}+2=-\frac{(a+b) (a+2 b)}{a^2+a b+b^2}$$ $$\frac{c}{b}+2=\frac{b (b-a)}{a^2+a b+b^2}$$  $$\frac{c}{a+b}-1=\frac{a (2 a+b)}{a^2+a b+b^2}$$
A: Thank you ,@Calvin Lin,
I  have solve this problem ,let $$F(a,b,c)=\left(\dfrac{b+c}{a}+2\right)^2+\left(\dfrac{c}{b}+2\right)^2+\left(\dfrac{c}{a+b}-1\right)^2$$ 
note 
$$F(a,b,c)-5=\dfrac{\left((a^2+ab+b^2)c+b(a+b)(2a+b)\right)^2}{a^2b^2(a+b)^2}\ge 0$$
