Let a, b, c be nonnegative reals with $a^ 2 + b^ 2 + c^ 2 + abc = 4$. Prove that $0 ≤ ab + bc + ca − abc ≤ 2$

Question

Let a, b, c be nonnegative reals with $a^ 2 + b^ 2 + c^ 2 + abc = 4$. Prove that $0 ≤ ab + bc + ca − abc ≤ 2$

We define $$U=[(a,b,c)|a,b,c>0; a^2+b^2+c^2<1000]$$This is intersection of open sets. Hence its open. It's closure is $$ \bar{U}=[a,b,c|a,b,c\geq{0};a^2+b^2+c^2 \leq{1000}] $$ Here the constraint set is $$\bar{S}=[x \in \bar{U}|g(x)=4]$$ is compact where $g(a,b,c)=a^2+b^2+c^2+abc$

Define $$f(a,b,c)=a^2+b^2+c^2+ab+bc+ca$$ It is equivalent to show that $f \leq 6$ subject to g. Over $\bar{S}$ it must achieve a global maximum.

If $x$ lies on the boundary that means one of its component is zero. My question is why is it so that one of the component must be zero?

Question Source: Lagrange Murderpliers Done Correctly by Evan Chen (page-5)

My Experience with calculus: I know what derivatives and integrals are and I know a few applications of derivatives and integrals.

Answer 1

The boundary here is the set $$S_1 = \{x\in \bar{U} - U | g(x) = 4\}.$$ So, it is just $a = 0, g(x) = 4$, or $b = 0, g(x) = 4$, or $c = 0, g(x) = 4$, or $a^2 + b^2 + c^2 = 1000, g(x) = 4$. The last one is impossible.