$x,y,z$ 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 $\space$ $x^4y^4z^4(x^3+y^3+z^3)≤3$  ?

Let's use the arithmetic-geometric inequality.
That gives us $(xyz)^{\frac{1}{4}}≤\frac{1}{4}(x+y+z)\leq\frac{3}{4}$
thus $(xyz)^4\leq(\frac{3}{4})^{16}$
Now all you have to prove is that $x^3+y^3+z^3\leq3*(\frac{4}{3})^{16}$
But $x^3+y^3+z^3\leq 3^3+3^3+3^3=81$ because $x\leq3$ and $y\leq3$ and $z\leq3$.
Thus you have to prove that $3^{15+4}\leq4^{16}$.
