# Find minimum of $4(a^3 + b^3 + c^3) + 15abc$ subject to $a + b + c = 2$

$$a$$, $$b$$ and $$c$$ are three sides of a triangle such that $$a + b + c = 2$$. Calculate the minimum value of $$\large P = 4(a^3 + b^3 + c^3) + 15abc$$

Every task asking for finding the minimum value of an expression containing the product of all of the variables scares me.

Here what I've done.

Using the AM-GM inequality and the Schur's inequality, we have that

$$a^3 + b^3 + c^3 \ge 3abc \implies P \ge \dfrac{9}{2}(a^3 + b^3 + c^3 + 3abc)$$

$$\ge \dfrac{9}{2}[ab(a + b) + bc(b + c) + ca(c + a)] = \dfrac{9}{2}[ab(2 - c) + bc(2 - a) + ca(2 - b)]$$

$$\ge \dfrac{9}{2}[2(ab + bc + ca) - 3abc] \ge \dfrac{27}{2}[2\sqrt[\frac{3}{2}]{abc} - abc]$$

Let $$abc = m \implies m \le \left(\dfrac{a + b + c}{3}\right)^3 = \dfrac{8}{27}$$

The problem becomes

Find the minimum value of $$P' = 2\sqrt[\frac{3}{2}]{m} - m$$ when $$0 < m \le \dfrac{8}{27}$$.

which is invalid because there isn't a minimum with the given condition.

• It must be $\implies P \color{red}{\le} \dfrac{9}{2}(a^3 + b^3 + c^3 + 3abc)$ Commented Mar 18, 2019 at 7:52

Let $$a=b=c=\frac{2}{3}$$. Thus, $$P=8.$$

We'll prove that it's a minimal value of $$P$$.

Indeed, we need to prove that $$\sum_{cyc}(4a^3+5abc)\geq(a+b+c)^3$$ or $$3\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0,$$ which is true by Schur.

I do not know if you had to use AM-GM but the problem is quite simple using pure algebra.

Considering $$P = 4(a^3 + b^3 + c^3) + 15abc \qquad \text{with} \qquad a+b+c=2$$ eliminate $$c$$ from the constaint to get $$P=3 a^2 (8-9 b)-3 a (b-2) (9 b-8)+8 (3 (b-2) b+4)$$ Now $$\frac{\partial P}{\partial a}=6 a (8-9 b)-3 (b-2) (9 b-8)=0 \implies a=\frac{2-b} 2$$ Reusing the constaint, this gives $$c=a$$ and then $$a=b=c=\frac 23$$.

Plug in $$P$$ and get the result.

Edit

Just as @KaiKoike commented, there is a mistake above $$\frac{\partial P}{\partial a}=-3 (9 b-8) (2 a+b-2)$$ $$\frac{\partial P}{\partial b}=-3 (9 a-8) ( a+2b-2)$$

So, we can have the solutions $$a=\frac 23 \qquad b=c=\frac 89\implies P=\frac{10720}{729}=14.7051$$ $$a=b=c=\frac 23 \implies P=\frac{10720}{729}=8$$

• From the equation $\partial P/\partial a=0$, it seems that you can only conclude that either $a=(2-b)/2$ or $b=8/9$. In addition, $a=(2-b)/2$ and $a=c$ do not necessarily imply $a=b=c$: you can have $(a,b,c)=(5/9,8/9,5/9)$.
– Kai
Commented Jun 18, 2022 at 15:32
• @KaiKoike. You are totally correct. I shall edit and quote you for sure. Commented Jun 18, 2022 at 15:40
• No problem! However, note that the candidates you would obtain with your approach are $(a,b,c)=(2/9,8/9,8/9),(5/9,5/9,8/9),(2/3,2/3,2/3)$ and their permutations. At these points except for $(a,b,c)=(2/3,2/3,2/3)$, the function $P=P(a,b,c)$ attains the same value approximately equal to $8.3$.
– Kai
Commented Jun 18, 2022 at 16:56
• I find this problem quite interesting as a problem for Calculus (although the proof by Michael Rozenberg using Schur's inequality is also elegant). The reason I say so is because the point $(a,b,c)=(2/9,8/9,8/9)$ is a local maximum of the function $P=P(a,b,c)$ subject to the constraint $a+b+c=2$ and the point $(a,b,c)=(5/9,5/9,8/9)$ is a non-extremal critical point with zero Hessian determinant.
– Kai
Commented Jun 18, 2022 at 17:13

This is particularly easy to prove by calculation. With the substitution $$a=u$$, $$b=u+v$$, $$c=u+v+w$$, we get $$4(a^3+b^3 + c^3)+15 a b c- (a+b+c)^3 = 3(u v^2 + u v w + u w^2 + 2 v w^2 + w^3)$$

$$\bf{Added:}$$ We have equality if and only if $$w=0$$ and $$u v=0$$, that is, if and only if $$a=b=c$$, or one of the $$a$$, $$b$$, $$c$$ is $$0$$ and the other two are equal.

First, using the identity $$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca),$$ we have $$a^3 + b^3 + c^3 - 3abc = 8 - 6(ab + bc + ca). \tag{1}$$

Second, using three degree Schur $$a^3 + b^3 + c^3 + 3abc \ge ab(a + b) + bc(b + c) + ca(c + a),$$ we have $$a^3 + b^3 + c^3 + 6abc \ge ab(a + b) + bc(b + c) + ca(c + a) + 3abc$$ or (using $$ab(a + b) + abc = ab(a + b + c)$$ etc.) $$a^3 + b^3 + c^3 + 6abc \ge (a + b + c)(ab + bc + ca) = 2(ab + bc + ca). \tag{2}$$

Third, using (1) and (2), we have \begin{align*} 4(a^3 + b^3 + c^3) + 15abc &= 3(a^3 + b^3 + c^3 + 6abc) + (a^3 + b^3 + c^3 - 3abc) \\ &\ge 3 \cdot 2(ab + bc + ca) + 8 - 6(ab + bc + ca)\\ &= 8. \end{align*}

Also, when $$a = b = c = 2/3$$, we have $$4(a^3 + b^3 + c^3) + 15abc = 8$$.

Thus, the minimum of $$4(a^3 + b^3 + c^3) + 15abc$$ is $$8$$.