Find minimum and maximum of $P=a+b+c$

Question

Let $a,b,c\ge 0$ such that $a^2+b^2+c^2+abc=4$. Find minimum and maximum of $$P=a+b+c$$

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+)Maximum: Let $x=\frac{2\sqrt{ab}}{\sqrt{\left(c+a\right)\left(c+b\right)}};y=\frac{2\sqrt{bc}}{\sqrt{\left(a+b\right)\left(a+c\right)}};z=\frac{2\sqrt{ca}}{\sqrt{\left(b+c\right)\left(b+a\right)}}$

And by AM-GM inequality i proved that $P\le 3$ when $a=b=c=1$

+)Minimum: I will prove $P\ge 2$ and the equality occurs when $a=b=0;c=2$ but failed. I tried substituting and uvw

Answer 1

With the maximal value you are right.

For $c=0$ we get a value $2$ and it'a a minimal value because $$2\sum_{cyc}\sqrt{\frac{ab}{(a+c)(b+c)}}\geq2$$ it's $$\sum_{cyc}\sqrt{ab(a+b)}\geq\sqrt{\prod_{cyc}(a+b)},$$ which is obvious after squaring of the both sides: $$\sum_{cyc}(a^2b+a^2c+2\sqrt{a^2bc(a+b)(a+c)})\geq\sum_{cyc}(a^2b+a^2c+\frac{2}{3}abc).$$

Answer 2

For the minimum, note $P^2 = 4+2(ab+bc+ca)-abc \geqslant 4+6(abc)^{2/3}-abc \geqslant 4$, as $abc \in [0, 1]$ from the obvious AM-GM $4 = a^2+b^2+c^2+abc \geqslant 4(abc)^{3/4}$ and $f(t)=6t^2-t^3$ is increasing in $t\in [0, 1]$.

Answer 3

Obviously, $a,b,c\in \left[0,2\right]$ so $abc\leqslant 8$. Thus $$abc = \sqrt[3]{abc}\cdot \sqrt[3]{(ab)(bc)(ca)} \leqslant 2\left(\frac{ab+bc+ca}{3}\right) \leqslant 2(ab+bc+ca).\tag{1}$$

Thus $$ 4=a^2+b^2+c^2+abc \leqslant (a+b+c)^2.$$ Equality occurs when we have equality in (1), which happens only when both sides are $0$.