<s>[INST] How prove this $x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$ Question:

let $x,y,z>0$ and such $xyz=1$, show that
  $$x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$$

My idea: use AM-GM inequality
$$x^3+x^3+1\ge 3x^2$$
$$y^3+y^3+1\ge 3y^2$$
$$z^3+z^3+1\ge 3z^2$$
so
$$2(x^3+y^3+z^3)+3\ge 3(x^2+y^2+z^2)$$ 
But this is not my inequality,so How prove it? I know this condition is very important.but how use this condition? and this inequality is stronger
 [/INST] This is hardly a pretty answer, but then again this inequality seems rather tight.
Consider the function $f(t)=e^{3t}-2e^{2t}+1+t$. It has the following properties:


*

*$f'(t)=e^{2t}(3e^t-4)+1$, so $f(t)$ is tangent to $y=0$ at $t=0$;

*$f''(t)=e^{2t}(9e^t-8)$, so $f(t)$ is concave down on $(-\infty,\ln\frac89)$ and concave up on $(\ln\frac89,\infty)$. This implies that $f(t)$ has at most one zero on $(-\infty,\ln\frac89)$ and no zeroes on $(\ln\frac89,0)$ and $(0,\infty)$. In other words, $f(t)$ changes sign at most once.

*We have $$\begin{align*}f\left(\ln\frac23\right)&=\frac{11}{27}-\ln\left(1+\frac12\right)\\
&>\frac{11}{27}-\left(\frac12-\frac12(\frac12)^2+\frac13(\frac12)^3-\frac14(\frac12)^4+\frac15(\frac12)^5\right)\\
&=\frac1{8640}>0,\end{align*}$$
so $f(t)\geq0$ on $[\ln\frac23,\infty)$.


Returning to the original question, let $xyz=1$, and assume $x\leq y\leq z$. We want to show that $\sum(x^3-2x^2+1)\geq0$.
Case 1: $x\leq\frac23$. Then we can easily check that $g(t)=t^3-2t^2+1$ is decreasing on $(0,\frac43)$ and increasing on $(\frac43,\infty)$. Hence
$$\begin{align*}\sum(x^3-2x^2+1)
&=g(x)+g(y)+g(z)\\
&\geq g\left(\frac23\right)+2g\left(\frac43\right)\\
&=\frac{11}{27}+2\left(\frac{-5}{27}\right)\\
&=\frac1{27}>0.\end{align*}$$
Case 2: $x>\frac23$. Set $x=e^a,y=e^b,z=e^c$. Then $a+b+c=0$, $a,b,c>\ln\frac23$, so
$$\sum(x^3-2x^2+1)=f(a)+f(b)+f(c)-(a+b+c)\geq0.$$
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