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.
 A: 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.
A: Unless you are challenging highschool-level math competition, I think you don't have to deal with purely algebraic method.
Here is my proof using diffrentiation.
I will prove the 'log'ed problem, which is enough. 
Now I will define new function f, for avoid some analytical conflict.
$$
f(w) = w^2ln(w) \,\,on \,w>0,\,\, 0 \,\,on \,w=0 \quad(which\, is \,diffrentiable)\\
$$
$$
'log'ed \,\, problem : f(x+y+z)+f(x)+f(y)+f(z) - f(x+y) - f(y+z) - f(x+z) \geq0
$$
You can check above problem is sufficient by exp() on each side.
Now I will define left side equation by g(x,y,z).
By partial diffrentiate g by x,
$$
h(w) = wln(w) \,\,on \,w>0,\,\, 0 \,\,on \,w=0 \\
\frac{\partial g}{\partial x} = 2[h(x+y+z)+h(x)-h(x+y) -h(x+z)]
$$
now you can check above partial diffrential  is positive or zero by dividing case.
(case 1: x=0, case 2: x>0. In each case , you will have to diffrentiate equation by x,y,z each and show what is happen.)
then we get 
$$
\frac{\partial g}{\partial x},\frac{\partial g}{\partial y},\frac{\partial g} {\partial z} \geq0 \,.
$$
since g(x,y,z) = 0 if one of variables is zero, following argument guarantees the problem.
$$
g(x,y,z) = g(x,y,0) + \int_{0}^{z} \frac{\partial g} {\partial z} dz\quad\geq0 .
$$
I think you can solve your second problem with this hint. Since some analytical details are not important in calculus class, I skipped several steps. 
Sorry for bad latex ( in fact this is my first)
