Find the minimum of :$P=a+b+c-ab-bc-ca$ soluLet $a,b,c$ be positive real numbers and $a+b+c+abc=4$.
We can rewrite the first equation as $a+b+c=4-abc.$ Then,
\begin{align*}
P&=a+b+c-ab-bc-ca\\&=(4-abc)-ab-bc-ca\\&=4-abc-ab-bc-ca-(a+b+c+1)+(a+b+c+1)\\&=4-(abc+ab+bc+ca+a+b+c+1)+a+b+c+1\\&=4-(a+1)(b+1)(c+1)+a+b+c+1\\&=5+a+b+c-(a+1)(b+1)(c+1)\\&=5+(4-abc)-(a+1)(b+1)(c+1)\\&=9-abc-(a+1)(b+1)(c+1).
\end{align*}It seems that both $abc$ and $(a+1)(b+1)(c+1)$ are maximized when $a=b=c,$ which would give $a=b=c=1$ and then $P=9-(1)(1)(1)-(2)(2)(2)=9-(1+8)=0.$ However, I'm not sure how to prove that statement.
 A: We will show that $a+b+c \geq ab+bc+ca$, from which it follows that the minimum is 0. This is attained when $a=b=c=1$ or permutations of $ (2,2,0)$.
The equality conditions suggest that Shur's inequality is involved, and we likely want to use
$$ a^3 + b^3 + c^3 + 3abc \geq \sum ab(a+b)$$.
Let's rewrite this in terms of $P=a+b+c$, $Q=ab+bc+ca$, and $R=abc$ which are the terms that appear in the given inequality and condition. Schur's inequality can be rewritten as:
$$P^3 -4PQ + 9 R \geq 0.$$ 
The given condition can be written as $P + R = 4$.

Proof that $P \geq Q$.
Proof by contradiction. Suppose not, then $ P < Q$. Then, 
$$ \begin{array} \text{9}R & \geq 4PQ - P^3  & \text{(Schur's)}  \\
 & > 4P^2 - P^3 & \text{(Contradiction assumption)} \\
& = P^2 (4-P) \\
& = P^2 R. & \text{(Given condition)} \\ \end{array}$$ 
So $9 > P^2 \Rightarrow 3 > P$.
Then, $R \leq \sqrt[3]{\frac{P}{3}} < 1$,
which contradicts $4 = P+R < 3 + 1 = 4$.
Hence $P \geq R$.

Note: I'm not happy with the structure of this proof. I wish there was a way to write it up more directly. 
A: We may suppose that $a\le b\le c$.
Supposing that $a\gt 1$ gives $4=a+b+c+abc\gt 4$, so we have $a\le 1$.
Supposing that $c\lt 1$ gives $4=a+b+c+abc\lt 4$, so we have $c\ge 1$.
Now using $$b=\frac{4-a-c}{1+ac}$$
we have
$$\begin{align}P&=a+b+c-ab-bc-ca
\\\\&=a+c-ac+(1-a-c)b
\\\\&=a+c-ac+(1-a-c)\cdot \frac{4-a-c}{1+ac}
\\\\&=\frac{(a+c-ac)(1+ac)+4-5(a+c)+(a+c)^2}{1+ac}
\\\\&=\frac{ac(a+c-ac)+(a+c)-ac+4-5(a+c)+(a+c)^2}{1+ac}
\\\\&=\frac{ac(a+c-ac-1)+4-4(a+c)+(a+c)^2}{1+ac}
\\\\&=\frac{ac(1-a)(c-1)+(a+c-2)^2}{1+ac}
\\\\&\ge 0\end{align}$$
$P$ equals $0$ when $a=b=c=1$, so the minimum of $P$ is $\color{red}0$.
A: This problem can be taken down with some heavy machinery. From the initial condition, obtain $c=\frac{4-a-b}{ab+1}$. We now want to minimize $$f(a,b)=a+b-ab+\left(\frac{a+b-4}{ab+1}\right)(a+b-1),$$ where $a$, $b$ are on the region bounded by the x and y axes, and the line $x+y\leq 4$. Since $f$ is continuous, it must have a minimum on this region.
If this minimum lies on a border, then one of $a$, $b$, $c$ must be zero. In this case, taking WLOG $c=0$, the initial condition becomes $a+b=4$, and our value to minimize, $a+b-ab$. By AM-GM, this last expression attains a minimum value of $0$ when $a=b=2$.
Otherwise, if the minimum lies inside the triangle, both partial derivatives at that point must be zero. Churning out these derivatives (which is not that hard, since $f$ is symmetric), and setting to zero, we find that either $$a=\frac{\sqrt{5}+1}{2}\text{ or }a=\frac{4-2b}{b^2+1},$$and viceversa for $b$. However, $f\left(\frac{\sqrt{5}+1}{2},b\right)=f\left(a,\frac{\sqrt{5}+1}{2}\right)=\frac{7-3\sqrt{5}}{2}>0$, which as we've already established, is not the minimum. This implies $$a=\frac{4-2b}{b^2+1},\,\,b=\frac{4-2a}{a^2+1}\Rightarrow$$ $$a=\frac{4-2\left(\frac{4-2a}{a^2+1}\right)}{\left(\frac{4-2a}{a^2+1}\right)^2+1}\Rightarrow$$ $$a^3-5a^2+3a+1=0\Rightarrow$$ $$(a-1)(a^2-4a-1)=0.$$ This all implies $a=b=c=1$, which also gives a value of $0$ on the expression to minimize.
In conclusion, the minimum is $0$, which is attained at $(a,b,c)=(0,2,2)$, $(2,0,2)$, $(2,2,0)$, $(1,1,1)$.
