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?
$x^{4}+y^{4}+z^{4}\geq x^{2}yz+xy^{2}z+xyz^{2}$
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?
$x^{4}+y^{4}+z^{4}\geq x^{2}yz+xy^{2}z+xyz^{2}$
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 $x,y,z$.)
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..): $\sum_{cyclic}(x^2 - y^2)^2 + \sum_{cyclic}x^2(y-z)^2 \ge 0$
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) $
$x^{4}+y^{4}+z^{4}\geq x^{2}yz+xy^{2}z+xyz^{2}$