$x,y,z$ 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 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!
