<s>[INST] 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.
 [/INST] 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. 
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