<s>[INST] Maximum value of $abc$ for $a+b+c=5$ and $a^2+b^2+c^2=11$ 
$a,b,c$ are three real numbers such that $a+b+c=5$ and $a^2+b^2+c^2=11$,
what's the maximum value of $abc$?

I thought of a way, $ab+bc+ca$ is not hard to find, $a,b,c$ satisfy the cubic equation $x^3 - 5 x^2 + 7 x - abc = 0$ , then use the discriminant of the cubic equation non-negative.
The discriminant of $x^3 + A x^2 + B x + C=0$ is
$A^2 B^2 - 4 B^3 - 4 A^3 C + 18 A B C - 27 C^2$
Is there an easier way?
 [/INST] From condition we get $ab+bc+ca=7.$ Using the Cauchy-Schwarz inequality, we have
$$11 \geqslant a^2 + \frac{(b+c)^2}{2} = a^2 + \frac{(5-a)^2}{2},$$
so
$$a^2 + \frac{(5-a)^2}{2} \leqslant 11 \Rightarrow \frac 13 \leqslant a \leqslant 3.$$
Similar we get $\frac 13 \leqslant b,\,c \leqslant 3.$ Therefore
$$(a-3)(b-3)(c-3) \leqslant 0,$$
equivalent to
$$abc \leqslant 27-9(a+b+c)+3(ab+bc+ca).$$
So $abc \leqslant  27-9 \cdot 5+3 \cdot 7 = 3.$
Equality occur when $a=b=1,\,c=3$ and any permution.
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