$x,y,z$ positive real numbers , $x+y+z=3$ $\implies x^4y^4z^4(x^3+y^3+z^3)≤3$

If $x,y,z$ are positive real numbers with $x+y+z=3$ then how to prove (without using calculus) that $\space$ $x^4y^4z^4(x^3+y^3+z^3)≤3$ ?

We have, $$(x+y+z)^3=x^3+y^3+z^3+3(x+y)(y+z)(x+z) \ge x^3+y^3+z^3+\frac{8}{3}(x+y+z)(xy+yz+xz) = x^3+y^3+z^3+(xy+yz+xz)+(xy+yz+xz)..[8 \text{times}] \ge 9\sqrt[9]{(x^3+y^3+z^3)(xy+yz+xz)^8}$$ Since $$(xy+yz+xz)^2 = x^2y^2+y^2z^2+x^2z^2+2xyz(x+y+z) \ge 3xyz(x+y+z)$$ we have, $$ 9\sqrt[9]{(x^3+y^3+z^3)(xy+yz+xz)^8} \ge 9\sqrt[9]{3^8(x^3+y^3+z^3)x^4y^4z^4}$$ And so we have $$ 3 \ge x^4y^4z^4(x^3+y^3+z^3) \Box $$

we can prove this follow $$3^{14}x^4y^4z^4(x^3+y^3+z^3)\le (x+y+z)^{15}$$ then assuming $x+y+z=1$,and denoting $t=3(xy+yz+xz),q=xyz$ $$\Longleftrightarrow 3^{14}q^4(1-t+3q)\le 1\Longleftrightarrow 1-3^{14}q^4(1-t)-3^{15}q^5\ge 0$$ since $$3xyz(x+y+z)\le (xy+yz+xz)^2\Longleftrightarrow q\le\dfrac{t^2}{3^3}$$ so $$1-3^{14}q^4(1-3p)-3^{15}q^5\ge 1-3^{14}\left(\dfrac{t^2}{3^3}\right)^4(1-t)-3^{15}\left(\dfrac{p^2}{3^3}\right)^5=1-9t^8(1-t)-t^{10}$$ and $$1-9t^8+9t^9-t^{10}=(1-t)(t^9+(1+t+t^2+\cdots+t^7-8t^8))=t^9(1-t)+(1-t)^2(1+2t+3t^2+4t^3+\cdots+8t^7)\ge 0$$ since $$t=3(xy+yz+xz)\le(xy+yz+xz)^2=1$$ By done!