<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] Plugging in $c=5-a-b$ into the quadratic and simplifying shows that
$$b^2+(a-5)b+(a^2-5a+7)=0.\tag{1}$$
Because $b+c=5-a$ we see that $b$ and $c$ are precisely the roots of this quadratic, so
$$abc=a(a^2-5a+7)=a^3-5a^2+7a.\tag{2}$$
Then the quadratic $(1)$ has two real roots so its discriminant is positive, i.e.
$$0\leq(a-5)^2-4(a^2-5a+7)=-3a^2+10a-3=-(3a-1)(a-3),$$
which shows that $\tfrac13\leq a\leq3$. Hence checking $(2)$ for extrema on the interval $[\tfrac13,3]$ shows that it is maximal if and only if $a\in\{1,3\}$. By symmetry $a,b,c\in\{1,3\}$ and because $a+b+c=5$ it follows that $(a,b,c)=(1,1,3)$ up to permutation, so the maximum value of $abc$ is $3$.
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