AM/GM inequalities I need some help to prove this inequality... I guess one can use Jensen's then AM/GM inequalities.
Let $x_1, x_2, x_3, x_4$ be non- negative real numbers such that
$x_1 x_2 x_3 x_4 =1$.
We want to show that
$$x_1^3 + x_2^3 + x_3^3 + x_4^3 \ge x_1+x_2+x_3+x_4,$$
and also
$$x_1^3 + x_2^3 + x_3^3 + x_4^3 \ge \frac1{x_1}+ \frac1{x_2} +\frac1{x_3}+\frac1{x_4}.$$
Since $x↦x^3$ is convex on R+ by Jensen's Inequality we have $x_1^3+x_2^3+x_3^3+x_4^3≥4^{-2}(x_1+x_2+x_3+x_4)^3$ then using AM/GM inequality and since $x_1x_2x_3x_4=1,$ we can show that $(x_1+x_2+x_3+x_4)^3≥4$
Many thanks for your help.
 A: 
$\displaystyle\prod_{i=1}^n x_i=1,x_i\in\mathbb R^+$.
Show that

*

*$(a)\displaystyle\sum_{i=0}^n x_i^3\ge \displaystyle\sum_{i=0}^nx_i$

*$(b)\displaystyle\sum_{i=0}^n x_i^3\ge \displaystyle\sum_{i=0}^n\frac{1}{x_i}$

$(a)$ Let $f(x)=x^3-x$. Since $\frac{d^2}{dx^2}f(x)=6x>0\ \forall\ x>0$, Jensen's inequality $$\Rightarrow \dfrac{\displaystyle\sum _{i=1}^n(x^3_i-x_i)}n\ge f\left(\frac{\displaystyle\sum x_i}n\right)\ge f\left(\left(\displaystyle\prod_{i=1}^n x_i\right)^{\frac1n}\right)=f(1)=0\tag{1}$$
since $f(x)$ is increasing after $x=1$.
PS: Draw graph of $f(x)$ to understand it better.
A: The first inequality.
We need to prove that:
$$\sum_{cyc}x_1^3\geq\sum_{cyc}x_1\sqrt{\prod_{cyc}x_1}$$ or
$$\sum_{cyc}x_1^3\geq\sum_{cyc}\sqrt{x_1^3x_2x_3x_4},$$which  is true by Muirhead because
$$(3,0,0,0)\succ(1.5,0.5,0.5,0.5).$$
We can use also AM-GM:
$$\sum_{cyc}x_1^3=\frac{1}{6}\sum_{cyc}(3x_1^3+x_2^3+x_3^3+x_4^3)\geq\frac{1}{6}\sum_{cyc}6\sqrt[6]{x_1^9x_2^3x_3^3x_4^3}=\sum_{cyc}x_1.$$
Also, the Tangent Line method works:
$$\sum_{cyc}(x_1^3-x_1)=\sum_{cyc}\left(x_1^3-x_1-2\ln{x_1}\right)\geq0.$$
Indeed, let $f(x)=x^3-x-2\ln{x},$ where $x>0$.
Thus, $$f'(x)=3x^2-1-\frac{2}{x}=\frac{(x-1)(3x^2+3x+2)}{x},$$ which gives $x_{min}=1$, $$f(x)\geq f(1)=0$$ and we are done!
The second inequality we can prove by similar ways.
