<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] 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 $8$.
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