$x,y,z$ 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 $$ 3 \ge x^4y^4z^4(x^3+y^3+z^3) \Box $$
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 $3^{15+4}\leq4^{16}$.