<s>[INST] Prove $0\le ab+bc+ca-abc\le2$ A question asks

Let $a,b,c$ be non-negative reals such that $$a^2+b^2+c^2+abc=4$$
Prove that$$0\le ab+bc+ca-abc\le2$$

The lower bound can be proven as follows:
If $a,b,c>1$, then $$a^2+b^2+c^2+abc>4$$
This implies that at least one of $a,b,c$ is less than or equal to $1$, and we can assume WLOG that $a\le1$. Then,
$$ab+bc+ca-abc=a(b+c)+bc(1-a)\ge0$$
For the upper bound, I had the idea of treating $a^2+b^2+c^2+abc=4$ as a quadratic in $a$ but don't see any way to proceed from there. Please only give me hints!
 [/INST] 
$a^2+b^2+c^2+abc=4, a, b, c:$ non-negative. Prove that $0\leq ab+bc+ca-abc \leq 4.$

\begin{align}
&a, b, c: \text{non-negative.} \\
& \text{You found that }a \leq 1 \text{ or } b \leq 1 \text{ or } c \leq 1.\\
\Rightarrow  \; & 1 \leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}. \\
\therefore \; &  abc \leq ab+bc+ca, 0 \leq ab+bc+ca-abc. \\
\ \\
& (a-b)^2+(b-c)^2+(c-a)^2 \geq 0. \\
\Rightarrow \; & a^2+b^2+c^2-ab-bc-ca \geq 0. \\
\Rightarrow \; & a^2+b^2+c^2+abc-ab-bc-ca-abc \geq 0. \\
\Rightarrow \; & 4 = a^2+b^2+c^2+abc \geq ab+bc+ca+abc \geq ab+bc+ca-abc.
\end{align}
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