Find minimum value of $(ab-2a+4)^2 + (bc-2b+4)^2+ (ca-2c+4)^2$ where $0 \leq a,b,c \leq2$ Find minimum value of $(ab-2a+4)^2 + (bc-2b+4)^2+ (ca-2c+4)^2$ where $0 \leq a,b,c \leq2$
For this kind of problems it's usually easy to guess that the minimum value is obtained at $a=b=c$ or when the answer is at the boundary.
When $a=b=c$ we can easily find minimum = 27. When we are at the boundary, we let $c=2$ we can see the expression is greater than or equal to 24. But this solution seems hacky and i'd like to find a better way.

For $(a,b.c)=(2,1,0)$ we'll get a value $24$.
We'll prove that it's a minimal value.
For $a=2$ by C-S we obtain:
$$\sum_{cyc}(ab-2a+4)^2=(2b)^2+(bc-2b+4)^2+16=$$
$$=\frac{1}{2}(1+1)((2b)^2+(bc-2b+4)^2)+16\geq\frac{1}{2}(2b+bc-2b+4)^2+16=$$
$$=\frac{1}{2}(bc+4)^2+16\geq8+16=24,$$ which says that it's enough to solve our problem for $\{a,b,c\}\subset[0,2).$
Now, let $a=\frac{2x}{x+1},$ $b=\frac{2y}{y+1}$ and $c=\frac{2z}{z+1},$ where $x$, $y$ and $z$ are non-negatives.
Thus, we need to prove that:
$$\sum_{cyc}\left(\frac{4xy}{(x+1)(y+1)}-\frac{4x}{x+1}+4\right)^2\geq24$$ or
$$2\sum_{cyc}\frac{(xy+y+1)^2}{(x+1)^2(y+1)^2}\geq3$$ or
$$\sum_{cyc}(x^2y^2z^2+2x^2y^2z+x^2y^2-2x^2y+2x^2z+x^2+2x+1)\geq0,$$ which is true because
$$\sum_{cyc}(x^2y^2z^2+2x^2y^2z+x^2y^2-2x^2y+2x^2z+x^2+2x+1)>$$
$$>\sum_{cyc}(x^2y^2-2x^2y+x^2)=\sum_{cyc}x^2(y-1)^2\geq0.$$
