Školjka

%V0 Vrijedi da je $\frac{a+1}{a(a+2)}=\frac{1}{a}-\frac{1}{a(a+2)}$, ali i $\frac{a+1}{a(a+2)}=\frac{1}{a+2}+\frac{1}{a(a+2)}$. Kad zbrojimo jednakosti, dobivamo $2\frac{a+1}{a(a+2)}=\frac{1}{a}+\frac{1}{a+2}$. Dakle $2\sum \limits_{cyc} \frac{a+1}{a(a+2)}=\sum \limits_{cyc}\frac{1}{a}+\sum \limits_{cyc}\frac{1}{a+2}$. Općenito po $AH$ vrijedi da je $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$. Primjenom toga gore se dobiva da je $2\sum\limits_{cyc}\frac{a+1}{a(a+2)} \geq \frac{9}{a+b+c} + \frac{9}{a+b+c+6} \geq \frac{9}{3} + \frac{9}{9} = 4$, odnosno $\sum \limits_{cyc}\frac{a+1}{a(a+2)} \geq 2$, što je traženi minimum.