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

Question

$a$, $b$ 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.

Answer 1

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.

Answer 2

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 $$a=\frac 23 \qquad b=c=\frac 89\implies P=\frac{10720}{729}=14.7051$$ $$a=b=c=\frac 23 \implies P=\frac{10720}{729}=8$$

Answer 3

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 $8$.

Answer 4

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.