Problem 1802

1. Initial Inequality and Simplification:
We start with the given inequality:
$$x^{24} + \sqrt[5]{y^{60} + z^{40}} \geq \left( x^4 y^3 + \frac{1}{3} y^2 z^2 + \frac{1}{9} x^3 z^3 \right)^2$$
We use the fact that:
$$\sqrt[5]{y^{60} + z^{40}} \geq \frac{1}{\sqrt[5]{16}} y^{12} + \frac{1}{\sqrt[5]{16}} z^8$$
Therefore, we need to prove:
$$x^{24} + \frac{1}{\sqrt[5]{16}} y^{12} + \frac{1}{\sqrt[5]{16}} z^8 \geq \left( x^4 y^3 + \frac{1}{3} y^2 z^2 + \frac{1}{9} x^3 z^3 \right)^2$$

2. Expanding the Right-Hand Side:
Expanding the square on the right-hand side, we get:
$$\left( x^4 y^3 + \frac{1}{3} y^2 z^2 + \frac{1}{9} x^3 z^3 \right)^2 = x^8 y^6 + \frac{1}{9} y^4 z^4 + \frac{1}{81} x^6 z^6 + \frac{2}{3} x^4 y^5 z^2 + \frac{2}{9} x^7 y^3 z^3 + \frac{2}{27} x^3 y^2 z^5$$

3. Applying AM-GM Inequality:
We will use the Arithmetic Mean-Geometric Mean (AM-GM) inequality to prove the required inequality. The AM-GM inequality states that for non-negative real numbers $a_1, a_2, \ldots, a_n$:
$$\frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \cdots a_n}$$
with equality if and only if $a_1 = a_2 = \cdots = a_n$.

4. Breaking Down the Terms:
We will break down the terms on the right-hand side and apply AM-GM to each group of terms.

(i) For $x^{24} + z^8 + z^8 + z^8$:
$$x^{24} + 3z^8 \geq 4 \sqrt[4]{x^{24} z^{24}} = 4 x^6 z^6$$
Dividing by 324:
$$\frac{1}{324} x^{24} + \frac{1}{108} z^8 \geq \frac{1}{81} x^6 z^6$$

(ii) For $y^{12} + z^8 + z^8$:
$$y^{12} + 2z^8 \geq 3 \sqrt[3]{y^{12} z^{16}} = 3 y^4 z^{\frac{16}{3}}$$
Since $z > \frac{3}{2}$, we have $z^{\frac{16}{3}} > z^4$, thus:
$$3 y^4 z^{\frac{16}{3}} > \frac{9}{2} y^4 z^4$$
Dividing by 81:
$$\frac{2}{81} y^{12} + \frac{4}{81} z^8 \geq \frac{1}{9} y^4 z^4$$

(iii) For $x^{24} + y^{12} + y^{12}$:
$$x^{24} + 2y^{12} \geq 3 \sqrt[3]{x^{24} y^{24}} = 3 x^8 y^8$$
Since $y > \frac{3}{2}$, we have $y^8 > y^6$, thus:
$$3 x^8 y^8 > \frac{27}{4} x^8 y^6$$
Dividing by 27:
$$\frac{4}{27} x^{24} + \frac{8}{27} y^{12} \geq x^8 y^6$$

(iv) For $x^{24} + y^{12} + y^{12} + z^8$:
$$x^{24} + 2y^{12} + z^8 \geq 4 \sqrt[4]{x^{24} y^{24} z^8} = 4 x^6 y^6 z^2$$
Since $y > \frac{3}{2}$, we have $y^6 > y^5$, thus:
$$4 x^6 y^6 z^2 > 9 x^4 y^5 z^2$$
Dividing by 27:
$$\frac{2}{27} x^{24} + \frac{4}{27} y^{12} + \frac{2}{27} z^8 \geq \frac{2}{3} x^4 y^5 z^2$$

(v) For $x^{24} + x^{24} + x^{24} + y^{12} + y^{12} + z^8 + z^8 + z^8$:
$$3x^{24} + 2y^{12} + 3z^8 \geq 8 \sqrt[8]{x^{72} y^{24} z^{24}} = 8 x^9 y^3 z^3$$
Since $x > \frac{3}{2}$, we have $x^9 > x^7$, thus:
$$8 x^9 y^3 z^3 > 18 x^7 y^3 z^3$$
Dividing by 27:
$$\frac{1}{27} x^{24} + \frac{2}{81} y^{12} + \frac{1}{27} z^8 \geq \frac{2}{9} x^7 y^3 z^3$$

(vi) For $x^{24} + y^{12} + z^8 + z^8 + z^8 + z^8$:
$$x^{24} + y^{12} + 4z^8 \geq 6 \sqrt[6]{x^{24} y^{12} z^{32}} = 6 x^4 y^2 z^{\frac{16}{3}}$$
Since $z > \frac{3}{2}$, we have $z^{\frac{16}{3}} > z^5$, thus:
$$6 x^4 y^2 z^{\frac{16}{3}} > 9 x^3 y^2 z^5$$
Dividing by 243:
$$\frac{2}{243} x^{24} + \frac{2}{243} y^{12} + \frac{8}{243} z^8 \geq \frac{2}{27} x^3 y^2 z^5$$

5. Summing Up the Inequalities:
Summing up all the inequalities obtained, we get:
$$x^{24} + \frac{1}{\sqrt[5]{16}} y^{12} + \frac{1}{\sqrt[5]{16}} z^8 \geq x^8 y^6 + \frac{1}{9} y^4 z^4 + \frac{1}{81} x^6 z^6 + \frac{2}{3} x^4 y^5 z^2 + \frac{2}{9} x^7 y^3 z^3 + \frac{2}{27} x^3 y^2 z^5$$