Maximum value of $abc$ for $a+b+c=5$ and $a^2+b^2+c^2=11$

Question

$a,b,c$ 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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $(a,b,c)=(1,1,3).$

Answer 4

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 $3$.