Vi er ikke alle så forskjellige, heldigvis... Problem E:17 053

          From GMB 11/2024 on page 650, proposed for 8th grade, author Constantin NICOLAU, Curtea_de_Argeș. In translation:

          "Let $x,\;y,\;z\;$ be nonzero real numbers, such that $x+y+z=24\;$ and

             $xy+yz+zx=192.\;$  Calculate $\frac{x^4}{y}+\frac{y^4}{z}+\frac{z^4}{x}.$"

ANSWER CiP

1 536     $(x=y=z=8)$

                    Solution CiP

          From $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$ we obtain

$$24^2=x^2+y^2+z^2+2 \cdot 192$$

$\Rightarrow\;\; x^2+y^2+z^2=576-384=192$ so

$$x^2+y^2+z^2=xy+yz+zx. \tag{1}$$

But (1)$\;\Leftrightarrow\;\;2\cdot x^2+2\cdot y^2+2\cdot z^2-2xy-2yz-2zx=0\;\Leftrightarrow$

$\Leftrightarrow\;\;(x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)=0\;\Leftrightarrow$

$\Leftrightarrow\;\;(x-y)^2+(y-z)^2+(z-x)^2=0\;$ and this is possible in real numbers (being a sum of non-negative terms) only if $x-y=0=y-z=z-x.$ Therefore $x=y=z$; moreover $x=y=z=\frac{x+y+z}{3}=\frac{24}{3}=8.$

     We then obtain $\frac{x^4}{y}+\frac{y^4}{z}+\frac{z^4}{x}=3\cdot x^3=3\cdot 8^3=1536.$

$\blacksquare$