Prove $0\le ab+bc+ca-abc\le2$

Question

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!

Answer 1

If one of $a,b,c$ is zero or two of them are equal, the inequality holds (almots) trivially.

Consider the function $ f(a,b,c) = ab+bc+ca-2abc $, with $a,b,c\ge 0, g(a,b,c) = 1$, where $g(a,b,c) = a^2 + b^2 + c^2 + abc$. The domain is a compact subset of $\mathbb{R}^3$ so $f$ must admit maximum. Assume the maximum is strictly larger than $2$, then it is taken in the interior of the domain, and $a,b,c$ are pairwisely distinct. Thus we can apply Lagrange's multiplier to get \begin{equation} \nabla f + \lambda \nabla g = 0. \end{equation} Notice that $\partial_a g = 2a + bc > 0$, so we have \begin{equation} \frac{a+c -ac}{2b+ac} = \frac{b+c-bc}{2a+bc}, \end{equation} or \begin{equation} (a-b)(2a+2b+2c-2ac-2bc+c^2) = 0 \end{equation} and two similar equations. Since we assumed $a,b,c$ are pairwisely distinct, we yield \begin{equation} a^2 - 2ac- 2ab = b^2 - 2bc -2ba \end{equation} or $a+b-2c = 0$. Then we have $a = b = c$ for sure, a contradiction.

Answer 2

$a^2+b^2+c^2+abc=4, a, b, c:$ 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}