Una solución:
Tenemos que [tex]x + y = -z[/tex], [tex]x + z = -y[/tex], [tex]y + z = -x[/tex], entonces
[tex]x² + y² = z² - 2xy[/tex], [tex]x² + z² = y² - 2xz[/tex], [tex]y² + z² = x² - 2yz[/tex], entonces
[tex]\frac{x² + y²}{x + y} + \frac{x² + z²}{x + z} + \frac{y² + z²}{y + z} = \frac{x² - 2xy}{-z} + \frac{y² - 2xz}{-y} + \frac{x² - 2yz}{-x} =[/tex]
[tex]\require{cancel}= \frac{xyzz - 2x²y² + xyzy - 2x²z² + xyzx - 2y²z²}{-xyz} =[/tex]
[tex]= \frac{xyz\cancelto{0}{(x + y + z)} }{-xyz} + \frac{2xy}{z} + \frac{2xz}{y} + \frac{2yz}{x} =[/tex]
[tex]= \frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} + \frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} =[/tex]
[tex]= \frac{x(y² + z²)}{yz} + \frac{y(x² + z²)}{xz} + \frac{z(x² + y²)}{xy}[/tex] =
[tex]= \frac{x³}{yz} + \frac{y³}{xz} + \frac{z³}{xy}[/tex][tex]-2\cancelto{0}{(x + y + z)} =[/tex]
[tex]= \frac{x³}{yz} + \frac{y³}{xz} + \frac{z³}{xy}[/tex]
