Let $a,b,c>0, a^2+b^2+c^2+abc=4$. Then prove that $0\leq ab+bc+ca-abc\leq 2$.
My idea is substitution. First, I'll prove that $cos^2\alpha+cos^2\beta+cos^2\gamma+2cos\alpha cos\beta cos\gamma=1$ where $\alpha+\beta+\gamma=\pi$.
$cos^2\alpha+cos^2\beta+cos^2\gamma+2cos\alpha cos\beta cos\gamma=\frac{3+cos2\alpha+cos2\beta+cos2\gamma}{2}+2cos\alpha cos\beta cos\gamma=\frac{3}{2}+cos(\alpha+\beta)cos(\alpha-\beta)-\frac{1}{2}+cos^2(\alpha+\beta)+2cos\alpha cos\beta cos\gamma=1+2cos(\alpha+\beta)cos\alpha cos\beta+2cos\alpha cos\beta cos\gamma=1$
Therefore we can substitute $a=2cos\alpha,b=2cos\beta,c=2cos\gamma$ with $\alpha,\beta,\gamma\in(0,\frac{\pi}{2})$. Then the inequality becomes $0\leq cos\alpha cos\beta+cos\beta cos\gamma+cos\gamma cos\alpha-2cos\alpha cos\beta cos\gamma\leq\frac{1}{2}$.
LHS is easy. If one of $cos\alpha,cos\beta,cos\gamma$ is equal to $0$, we may assume that $\alpha=\frac{\pi}{2}$. $0\leq\beta,\gamma\leq\frac{\pi}{2}$ and $cos\beta cos\gamma\geq 0$(True).
Else, we can divide by $cos\alpha cos\beta cos\gamma$ and the inequality becomes $\frac{1}{cos\alpha}+\frac{1}{cos\beta}+\frac{1}{cos\gamma}\geq 2$. But it's true because $\frac{1}{cos\alpha}+\frac{1}{cos\beta}+\frac{1}{cos\gamma}\geq\frac{9}{cos\alpha+cos\beta+cos\gamma}\geq\frac{9}{\frac{3}{2}}=6>2$. Equality occurs if $(\alpha,\beta,\gamma)=(\frac{\pi}{2},\frac{\pi}{2},0)$ and its permutations.
I can't solve RHS. Maybe dividing by $cos\alpha cos\beta cos\gamma$, it can be little bit easier. Anyone can help me?
