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?
Yes, we need to find $ab+ac+bc=7$ before.
Now, $a$, $b$ and $c$ are roots of the equation:
$$(x-a)(x-b)(x-c)=0$$ or
$$abc=x^3-5x^2+7x.$$
Now, $$(x^3-5x^2+7x)'=(x-1)(3x-7),$$ which gives that a maximal value of $abc$ for which the equation $$abc=x^3-5x^2+7x$$ has three real roots holds for $x=1$, which gives:
$$\max_{a+b+c=7,a^2+b^2+c^2=11}{abc}=3.$$
The equality occurs for example, for $(a,b,c)=(1,1,3).$
As you did not exactly clarify what you mean by abc (their product?) I can only give you halve an answer. I assume you do not want a integer solution rather than a continous one and also real solutions.
With some calculations you will find:
$$b_{1/2}=\frac{1}{2}(+/- \sqrt{-3a^2+10a-3}-a+5) $$
and
$$c_{1/2}=\frac{1}{2}(+/- \sqrt{-3a^2+10a-3}-a+5) $$
This means you can define $f_{1/1}(x\in \mathbb{R}):=a*b_1*c_1$, $f_{1/2}(x\in \mathbb{R}):=a*b_1*c_2$ and so on (all combinations) and from them just calculate the maximum by using standard analysis methods.