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.

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 $\color{red}0$.
