First, note that $\sqrt[5]{y^{60}+z^{40}} \geq \frac{1}{\sqrt[5]{16}}y^{12}+\frac{1}{\sqrt[5]{16}}z^8$. So, we have to prove that $x^{24}+\frac{1}{\sqrt[5]{16}}y^{12}+\frac{1}{\sqrt[5]{16}}z^8 \geq x^8y^6+\frac{1}{9}y^4z^4+\frac{1}{81}x^6z^6+\frac{2}{3}x^4y^5z^2+\frac{2}{9}x^7y^3z^3+\frac{2}{27}x^3y^2z^5.$ Next, I will use a bunch of A.M-G.M: (i) $x^{24}+z^8+z^8+z^8 \geq 4x^6z^6$, so $$\frac{1}{324}x^{24}+\frac{1}{108}z^8 \geq \frac{1}{81}x^6z^6.$$(ii) $y^{12}+z^8+z^8 \geq 3y^4z^{\frac{16}{3}}>\frac{9}{2}y^4z^4$, so $$\frac{2}{81}y^{12}+\frac{4}{81}z^8 \geq \frac{1}{9}y^4z^4.$$(iii) $x^{24}+y^{12}+y^{12} \geq 3x^8y^8>\frac{27}{4}x^8y^6$, so $$\frac{4}{27}x^{24}+\frac{8}{27}y^{12} \geq x^8y^6.$$(iv) $x^{24}+y^{12}+y^{12}+z^8 \geq 4x^6y^6z^2>9x^4y^5z^2$, so $$\frac{2}{27}x^{24}+\frac{4}{27}y^{12}+\frac{2}{27}z^8 \geq \frac{2}{3}x^4y^5z^2.$$(v) $x^{24}+x^{24}+x^{24}+y^{12}+y^{12}+z^8+z^8+z^8 \geq 8x^9y^3z^3 >18x^7y^3z^3.$, so $$\frac{1}{27}x^{24}+\frac{2}{81}y^{12}+\frac{1}{27}z^8 \geq \frac{2}{9}x^7y^3z^3.$$(vi) $x^{24}+y^{12}+z^8+z^8+z^8+z^8 \geq 6x^4y^2z^{\frac{16}{3}}>9x^3y^2z^5$, so $$\frac{2}{243}x^{24}+\frac{2}{243}y^{12}+\frac{8}{243}z^8 \geq \frac{2}{27}x^3y^2z^5.$$Summing up all 6 inequalities above, we get the desired result. Note. We need to check that $\frac{1}{324}+\frac{4}{27}+\frac{2}{27}+\frac{1}{27}+\frac{2}{243}\leq1$, $\frac{2}{81}+\frac{8}{27}+\frac{4}{27}+\frac{2}{81}+\frac{2}{243}\leq\frac{1}{\sqrt[5]{16}}$, and $\frac{1}{108}+\frac{4}{81}+\frac{2}{27}+\frac{1}{27}+\frac{8}{243}\leq\frac{1}{\sqrt[5]{16}}$ which are not hard to prove.