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$a$, $b$ 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.

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  • $\begingroup$ It must be $\implies P \color{red}{\le} \dfrac{9}{2}(a^3 + b^3 + c^3 + 3abc)$ $\endgroup$ – farruhota Mar 18 '19 at 7:52
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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.

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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.

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