$x,y,z$ 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 $\space$ $x^4y^4z^4(x^3+y^3+z^3)≤3$  ?
 A: 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 $$ 3 \ge x^4y^4z^4(x^3+y^3+z^3) \Box $$
A: 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!
A: Well, I tried and failed to find an answer made of sugar and spice. Here is a frogs and snails answer.
As karafka noted, the AM-GM inequality easily implies that $xyz\leq 1$. So let $\epsilon\geq 0$ be such that $xyz=1-\epsilon$ . 
Let us order the variables so that $x\leq y\leq z$. We must have $z\geq 1$ and $x\leq 1$.
Now, it can be checked that \begin{eqnarray*} (xyz)^4(x^3+y^3+z^3) & = & (x+y+z)^3(xyz)^4-3(x+y+z)(xy+xz+yz)(xyz)^4+3(xyz)^5,\end{eqnarray*}
and since $x+y+z=3$ we deduce
\begin{eqnarray*} (xyz)^4(x^3+y^3+z^3) & = & 27(1-\epsilon)^4-9(xy+xz+yz)(1-\epsilon)^4+3(1-\epsilon)^5\\ &=& 3\left(10-\epsilon+3(xy+xz+yz)\right)(1-\epsilon)^4\end{eqnarray*}
Thus we're done if we can show that $\left(10-\epsilon-3(xy+xz+yz)\right)\leq \frac{1}{(1-\epsilon)^4}.$
Now, since $$1+\epsilon+\epsilon^2 = \frac{1-\epsilon^3}{1-\epsilon}\leq \frac{1}{1-\epsilon},$$ we can take the fourth power of both sides. Then note that $$1+4\epsilon\leq 1+4\epsilon+10\epsilon^2\leq \frac{1}{(1-\epsilon)^4}.$$
Case 1: $\frac{3}{5}\leq x \leq 1$.
In this case we aim to show that $\left(10-\epsilon-3(xy+xz+yz)\right)\leq 1+4\epsilon$, i.e, that $C\geq 0$, where $C=1+4\epsilon - \left(10-\epsilon-3(xy+xz+yz)\right)$.
Since $\frac{3}{5}\leq x\leq 1$ and $z\geq 1$ there exists $a\in[0,\frac{2}{5}]$ such that $x=1-a$ and $b\in[0,2)$ such that $1+b=z$. Hence $y=1+a-b$. Fix such an $a$. 
Now, $C$ can be written in terms of $a$ and $b$ as $C=(2-5a)b^2+(5a^2-2a)b+2a^2$. Since $a\leq \frac{2 }{5}$, this corresponds to an upwards parabola in the $C-b$ plane with vertex $\left(\frac{a}{2}, \frac{a^2}{5}(6+5a)\right)$. Hence $C\geq0$ for every choice of $b$.
Case 2: $\frac{3}{25}\leq x\leq \frac{3}{5}$.
In this case, we'll show that $\left(10-\epsilon-3(xy+xz+yz)\right)\leq 1+4\epsilon+10\epsilon^2$. For this case, write the right side minus the left in terms of $x,y,z$ to obtain:
$$C=10x^2(yz)^2+(3-25x)(yz)+6+9x-3x^2.$$
Now fix an $x$ in the range. Then $C$ corresponds to upward parabola in "variable" $yz$. Its discriminant is $120x^4-360x^3+385x^2-150x+9.$ Using the quartic formula (...) , this polynomial has real roots $r_1\approx 0.08<\frac{3}{25}$ and $r_2\approx 0.78>\frac{3}{5}$, and is negative in between. Thus for any valid choice of $y,z$, we must have $C>0$.
Case 3: $0<x<\frac{3}{25}$.
The vertex of the parabola from Case $2$ is located at $yz=\frac{25x-3}{20x^2}$. However, this is negative since $x<\frac{3}{25}$. Since the parabola is upwards with $C$-intercept at $C=6+9x-3x^2>6-3>0$, this means that $C>0$ when $yz$ is positive (which it is).
A: What about this idea: the function is convex on the domain $x,y,z>0$. The constraint $x+y+z=3$ is a convex subdomain. Therefore the function is convex on this subdomain. It can have only one local maximum inside this region, and the symmetry dictates that it must be at $x=y=z=1$.
The maximum of this function can only occur either on the boundary or at the specified point. At the boundary, it equals zero, so the only other candidate for a maximum is
$1^41^41^4(1^3+1^3+1^3)=3$
A: 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 $3^{15+4}\leq4^{16}$.
