# Prove $x^{4}+y^{4}+z^{4}\geq x^{2}yz+xy^{2}z+xyz^{2}$ trivially?

I need to prove several inequalities trivially. (i.e. without using AM-GM, re-arrangement etc). I just keep hitting a blank. Could anyone help?

$$x^{4}+y^{4}+z^{4}\geq x^{2}yz+xy^{2}z+xyz^{2}$$

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$(x,y,z)=(1,1,1)$ – pedja Apr 8 '12 at 10:36
@pedja: There are other solutions, e.g., $\langle x,y,z\rangle=\langle -1,1,1\rangle$. – Brian M. Scott Apr 8 '12 at 10:36
Sorry I forgot to add that I have to prove this inequality holds true for all positive real numbers x,y,z. – Roy Apr 8 '12 at 11:05
Using Muirhead's inequality these can be easily proven. – karakfa Dec 21 '12 at 19:44

The righthand side is clearly $xyz(x+y+z)$. Assume that $x,y,z>0$. For fixed $x+y+z$, $xyz$ is maximized when $x=y=z$, and $x^4+y^4+z^4$ is minimized when $x=y=z$. This is easy to see if you can visualize the surfaces $x+y+z=k$, $xyz=k$, and $x^4+y^4+z^4=k$ for a positive constant $k$. Thus, for a fixed value of $x+y+z$, the worst case for the inequality is when $x=y=z$: that’s when the righthand side is biggest and the lefthand side is smallest. But when $x=y=z$, the two sides are clearly equal, so the inequality holds for all $x,y,z>0$. (From there, by the way, it’s easy to show that it holds for all real $x,y,z$.)

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I'm guessing Sum-of-Squares representation would amount to a 'trivial' proof, right? How about expanding the following (this further proves that your inequality holds for all reals, not just positive ones..): $$\sum_{cyclic}(x^2 - y^2)^2 + \sum_{cyclic}x^2(y-z)^2 \ge 0$$

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we know that, $$\frac{x^{4}+y^{4}+z^{4}}{3}\geq (\frac{x+y+z}{3})^{4}$$

$$x^{4}+y^{4}+z^{4}\geq \frac{1}{27}(x+y+z)^{4}$$

$\frac{x+y+z}{3} \geq (xyz)^{\frac{1}{3}}$

$(x+y+z)^3 \geq 27xyz$

hence $$x^{4}+y^{4}+z^{4}\geq \frac{1}{27}(x+y+z)(27xyz)$$

$$x^{4}+y^{4}+z^{4}\geq x^{2}yz+xy^{2}z+xyz^{2}$$

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