5. Zadatak

Neka su $a$ , $b$ i $c$ racionalni brojevi takvi da je

\frac{a}{b(a+b)}+\frac{b}{c(b+c)}+\frac{c}{a(c+a)}=2.

Koliko tada iznosi

\frac{b}{a(a+b)}+\frac{c}{b(b+c)}+\frac{a}{c(c+a)}?

Prvo rješenje.

Redom računamo

\begin{aligned} \frac{b}{a(a+b)}+\frac{c}{b(b+c)}+\frac{a}{c(c+a)} &=\frac{b+a-a}{a(a+b)}+\frac{c+b-b}{b(b+c)}+\frac{a+c-c}{c(c+a)}\\ &=\frac{b+a}{a(a+b)}-\frac{a}{a(a+b)}+\frac{c+b}{b(b+c)}-\frac{b}{b(b+c)}+\frac{a+c}{c(c+a)}-\frac{c}{c(c+a)}\\ &=\frac{1}{a}-\frac{1}{a+b}+\frac{1}{b}-\frac{1}{b+c}+\frac{1}{c}-\frac{1}{c+a}\\ &=\left(\frac{1}{b}-\frac{1}{a+b}\right)+\left(\frac{1}{c}-\frac{1}{b+c}\right)+\left(\frac{1}{a}-\frac{1}{c+a}\right)\\ &=\frac{a+b-b}{b(a+b)}+\frac{b+c-c}{c(b+c)}+\frac{c+a-a}{a(c+a)}\\ &=\frac{a}{b(a+b)}+\frac{b}{c(b+c)}+\frac{c}{a(c+a)}=2. \end{aligned}

Drugo rješenje.

Vrijedi

\begin{aligned} &\frac{a}{b(a+b)}+\frac{b}{c(b+c)}+\frac{c}{a(c+a)} -\left(\frac{b}{a(a+b)}+\frac{c}{b(b+c)}+\frac{a}{c(c+a)}\right)\\ &=\frac{a}{b(a+b)}-\frac{b}{a(a+b)}+\frac{b}{c(b+c)}-\frac{c}{b(b+c)}+\frac{c}{a(c+a)}-\frac{a}{c(c+a)}\\ &=\frac{a^2-b^2}{ab(a+b)}+\frac{b^2-c^2}{bc(b+c)}+\frac{c^2-a^2}{ca(c+a)}. \end{aligned}

\begin{aligned} a^2-b^2&=(a-b)(a+b),\\ b^2-c^2&=(b-c)(b+c),\\ c^2-a^2&=(c-a)(c+a) \end{aligned}

slijedi

\begin{aligned} \frac{a^2-b^2}{ab(a+b)}+\frac{b^2-c^2}{bc(b+c)}+\frac{c^2-a^2}{ca(c+a)} &=\frac{(a-b)(a+b)}{ab(a+b)}+\frac{(b-c)(b+c)}{bc(b+c)}+\frac{(c-a)(c+a)}{ca(c+a)}\\ &=\frac{a-b}{ab}+\frac{b-c}{bc}+\frac{c-a}{ca}\\ &=\frac{1}{b}-\frac{1}{a}+\frac{1}{c}-\frac{1}{b}+\frac{1}{a}-\frac{1}{c}=0. \end{aligned}

Dakle,

\frac{b}{a(a+b)}+\frac{c}{b(b+c)}+\frac{a}{c(c+a)} =\frac{a}{b(a+b)}+\frac{b}{c(b+c)}+\frac{c}{a(c+a)}=2.