# Find the smallest integer $n$ such that $(x^2 + y^2 + z^2)^2 \leq n(x^4 + y^4 + z^4)$ for all real numbers $x, y,$ and $z.$

Find the smallest integer $n$ such that

$$(x^2 + y^2 + z^2)^2 \leq n(x^4 + y^4 + z^4)$$for all real numbers $x, y,$ and $z.$

How should I manipulate this inequality? I am stuck and don't know how to proceed. All solutions are greatly appreciated!

• This smells of Cauchy-Schwarz... I'm not quite sure. – Sean Roberson Sep 24 '16 at 21:22

Let's work with $a=x^2$, $b=y^2$, and $c=z^2$ which are all nonnegative. Then $$(a+b+c)^2\leq 3(a^2+b^2+c^2) \tag{*}$$ (either by Cauchy-Schwarz or by expanding both sides) so $n\leq 3$. But ($*$) is an equality when $a=b=c>0$ so $n\geq 3$. We infer that $n=3$.
$$(x^2 + y^2 + z^2)^2 \leq n(x^4 + y^4 + z^4)\iff (xy)^2+(xz)^2+(yz)^2\le \frac {n-1}{2}(x^4+y^4+z^4)$$ If $n=3-\epsilon$ where $\epsilon\gt 0$ and $x=y=z$ then $2\le 2-\epsilon$, absurd. Thus the smallest $n$ is $3$.