How prove this $x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$


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
$$x^3+x^3+1\ge 3x^2$$
$$y^3+y^3+1\ge 3y^2$$
$$z^3+z^3+1\ge 3z^2$$
$$2(x^3+y^3+z^3)+3\ge 3(x^2+y^2+z^2)$$

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

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Here is a possible solution: (although it is not the most elegant one)

I will employ Mixing Variables technique here. Since the inequality is symmetric, WLOG let $x=\min(x,y,z)$. Therefore $t^2:=yz \ge 1$.
I wish to show
$$f(x,y,z)\ge f(x,\sqrt{yz},\sqrt{yz}) = f(\frac1{t^2},t,t) \ge -3$$
Let us put $p^2=x, q^2=y, r^2=z$. The first inequality in the above chain is equivalent to
$$q^6+r^6-2q^3r^3 \ge 2(q^4+r^4-2q^2r^2)$$
$$\iff (q^3-r^3)^2 \ge 2(q^2-r^2)^2$$
$$\iff (q^2+qr+r^2)^2 \ge 2(q+r)^2$$
This is true since
$$(q^2+qr+r^2)^2 \ge q^4+r^4+2q^2r^2+2qr(q^2+r^2) \ge 4q^2r^2+2(q^2+r^2) \ge 2(q+r)^2$$
Therefore it enough to prove $f(\frac1{t^2},t,t)\ge -3$ for $t>0$ which is equivalent to
$$(t-1)^2((t^7-2t^5+t^3)+(t^7+t-2t^4)+(t^4-t^3+t^2)+t+1)\ge 0$$
Each term in the brackets of the large factor is greater than zero by AM-GM.

The last part is little tedious to do by hand. But you always know that $(t-1)$ has to be factor (possibly with multiplicity $2$) of that thing. That helps in simplification.

Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, our inequality is equivalent to $f(v^2)\geq0$, where $f$ is a linear function.
Hence, $f$ gets a minimal value, when $v^2$ gets an extremal value, which happens when
two numbers from $\{x,y,z\}$ are equal.
Id est, it remains to prove our inequality for $y=x$ and $z=\frac{1}{x^2}$, which gives something obvious.

I have been trying to find a more aesthetic answer, but until I do, I will post a variational approach.

I have simplified the argument a bit, but I am still looking for a simpler approach.

To minimize $x^3+y^3+z^3-2x^2-2y^2-2z^2$ given $xyz=1$, we need to find $x,y,z$ so that for all variations $\delta x,\delta y,\delta z$ that maintain $xyz=1$; that is,
\frac{\delta x}{x}+\frac{\delta y}{y}+\frac{\delta z}{z}=0\tag{1}
we have $x^3+y^3+z^3-2x^2-2y^2-2z^2$ is stationary:
(3x^2-4x)\,\delta x+(3y^2-4y)\,\delta y+(3z^2-4z)\,\delta z=0\tag{2}
Standard orthogonality arguments imply that there is a $\lambda$ so that $x,y,z$ satisfy
Since $3t^3-4t^2$ decreases on $\left[0,\frac89\right]$ and increases for $t\gt\frac89$, for any value of $\lambda$, there can be at most two positive values for $x,y,z$; thus, two must be the same. Say $y=x$ and $z=x^{-2}$. Both $x$ and $x^{-2}$ must satisfy $(3)$, therefore,
Using Sturm’s Theorem, we see that $3x^9-4x^8+4x^2-3$ has only one real root; that is $x=1$. Plugging this into the expression to be minimized gives a minimum of $-3$. This means

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)\\
    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
&\geq g\left(\frac23\right)+2g\left(\frac43\right)\\

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