<s>[INST] 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.
 [/INST] 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.
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