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