3. Zadatak

Prvo rješenje.

Pomnožimo li uvjet s $x+y+z+w$, dobivamo

$$\frac{x^2+x(y+z+w)}{y+z+w}+\frac{y^2+y(x+z+w)}{z+w+x}+\frac{z^2+z(x+y+w)}{w+x+y}+\frac{w^2+w(x+y+z)}{x+y+z}=x+y+z+w.$$

tj.

$$\frac{x^2}{y+z+w}+x+\frac{y^2}{z+w+x}+y+\frac{z^2}{w+x+y}+z+\frac{w^2}{x+y+z}+w=x+y+z+w.$$

Odavde slijedi

$$\frac{x^2}{y+z+w}+\frac{y^2}{z+w+x}+\frac{z^2}{w+x+y}+\frac{w^2}{x+y+z}=0.$$

Drugo rješenje.

Označimo $s=x+y+z+w$. Tada je

$$\frac{x}{y+z+w}=\frac{s-(y+z+w)}{y+z+w}=\frac{s}{s-x}-1,$$

i analogno za ostale pribrojnike. Početni uvjet dakle glasi

$$\frac{s}{s-x}+\frac{s}{s-y}+\frac{s}{s-z}+\frac{s}{s-w}=5.$$

Također možemo zapisati

$$\frac{x^2}{y+z+w}=\frac{x^2-s^2}{s-x}+\frac{s^2}{s-x}=\frac{(x-s)(s+x)}{s-x}+s\cdot\frac{s}{s-x}=-(s+x)+s\cdot\frac{s}{s-x},$$

i analogno za ostale pribrojnike. Traženi zbroj iznosi

$$\begin{aligned} &\frac{x^2}{y+z+w}+\frac{y^2}{z+w+x}+\frac{z^2}{w+x+y}+\frac{w^2}{x+y+z}\\ &=s\left(\frac{s}{s-x}+\frac{s}{s-y}+\frac{s}{s-z}+\frac{s}{s-w}\right)\\ &\qquad -(s+x)-(s+y)-(s+z)-(s+w)\\ &=5s-4s-(x+y+z+w)=s-s=0. \end{aligned}$$

Treće rješenje.

Prebacimo li sve osim prvog pribrojnika u uvjetu na drugu stranu i pomnožimo s $x$, dobivamo

$$\frac{x^2}{y+z+w}=x-\frac{xy}{z+w+x}-\frac{xz}{w+x+y}-\frac{xw}{x+y+z}.$$

Analogno, ostavljanjem svakog od preostala tri pribrojnika iz uvjeta s jedne strane i množenjem s $y$, $z$, $w$ redom, dobivamo jednadžbe

$$\frac{y^2}{z+w+x}=y-\frac{xy}{y+z+w}-\frac{yz}{w+x+y}-\frac{yw}{x+y+z},$$ $$\frac{z^2}{w+x+y}=z-\frac{xz}{y+z+w}-\frac{yz}{z+w+x}-\frac{zw}{x+y+z},$$ $$\frac{w^2}{w+x+y}=w-\frac{xw}{y+z+w}-\frac{yw}{z+w+x}-\frac{zw}{w+x+y}.$$

Zbrajanjem te četiri jednadžbe dobivamo

$$\begin{aligned} &\frac{x^2}{y+z+w}+\frac{y^2}{z+w+x}+\frac{z^2}{w+x+y}+\frac{w^2}{x+y+z}\\ &=x+y+z+w-\frac{x(y+z+w)}{y+z+w}-\frac{y(z+w+x)}{z+w+x}-\frac{z(w+x+y)}{w+x+y}-\frac{w(x+y+z)}{x+y+z}\\ &=x+y+z+w-x-y-z-w=0. \end{aligned}$$