How to prove $(x+y+z)^{x+y+z}x^xy^yz^z \le (x+y)^{x+y}(y+z)^{y+z}(z+x)^{z+x}$ where $x, y, z \ge 0$? 
For $x, y, z \in \mathbb R^+\cup\{0\} $, prove that
  $$(x+y+z)^{x+y+z}x^xy^yz^z \le (x+y)^{x+y}(y+z)^{y+z}(z+x)^{z+x}$$ and
  $$(x+y+z)^{(x+y+z)^2}x^{x^2}y^{y^2}z^{z^2} \ge (x+y)^{(x+y)^2}(y+z)^{(y+z)^2}(z+x)^{(z+x)^2}$$

[Edited] I currently solved the first one. Please check the procedures I made:

$$x(x+y+z)=x^2+xy+xz \le x^2+xy+xz+yz=(x+y)(z+x) $$
$$x^x(x+y+z)^{x} \le (x+y)^x(z+x)^x \cdots (1)$$ In the same way,
  $$y^y(x+y+z)^y \le (x+y)^y(y+z)^y \cdots (2)$$
$$z^z(x+y+z)^z \le (y+z)^z(z+x)^z \cdots (3)$$
  When we multiply all the three inequalities (1)~(3), since $x, y, z\ge0$:
  $$(x+y+z)^{x+y+z}x^xy^yz^z \le (x+y)^{x+y}(y+z)^{y+z}(z+x)^{z+x}$$

However, I do not have any ideas to prove the second inequality although I spent about a whole week to solve this. I'm very confused since the direction of the sign of the second inequality is opposite to the previous one. 
Also, this question belongs to the calculus. How can I apply the calculus in the proof? Thanks for your help.

It should be $x>0$, $y>0$ and $z>0$, otherwise $x^x$ is not defined for $x=0$.
Let $f(z)=\sum\limits_{cyc}(x+y)\ln(x+y)-\sum\limits_{cyc}x\ln{x}-(x+y+z)\ln(x+y+z).$
Thus, $$f'(z)=\ln(x+z)+1+\ln(y+z)+1-\ln{z}-1-\ln(x+y+z)-1=$$
$$=\ln\frac{(x+z)(y+z)}{z(x+y+z)}=\ln\left(1+\frac{xy}{z(x+y+z)}\right)>0.$$
Thus, $$f(z)\geq \lim_{z\rightarrow0^+}f(z)=0$$ and we are done!
The second problem we can solve by the same way.
