<s>[INST] Find $a,b,c \in \mathbb{R^+}$ such that $a^2+b^2+c^2+abc=4$ and $a+b+c=3$ 
Find  $a,b,c \in \mathbb{R^+}$ such that $$a^2+b^2+c^2+abc=4\;\;\;{\rm  and}\;\;\;a+b+c=3$$

Seeing $a^2+b^2+c^2+abc=4$,I substituted $a=2\cos{A}$,$b=2\cos{B},c=\cos{C}$ with $A+B+C=180°$.
Therefore,$cosA+cosB+cosC=\frac 32$ have thousands of solution. What to do now? Please help me.
 [/INST] If we put $c=3-a-b$ we get quadratic equation on $a$ with parameter $b$, if $b \ne 2$:
$$(2-b)a^2-a(b^2-5b+6)+(2b^2-6b+5)=0$$
which has a solution iff it discriminat is nonnegative: $$(b-2)(b+2)(b-1)^2\geq 0$$
so $b>2$ or $b=1$...


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*Clearly if $b=1$ we get $a^2-2a+1=0$ so $a=1$ and then $c=1$.

*If $b>2$ then with similar reason say on $b$ we get $a>2$ or $c>2$ which is impossibile.

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