4. Zadatak

Prvo rješenje.

Primjenom A–H nejednakosti dobivamo

$$\begin{aligned} \frac{a+1}{a(a+2)}+\frac{b+1}{b(b+2)}+\frac{c+1}{c(c+2)} &\ge \frac{9}{\frac{a(a+2)}{a+1}+\frac{b(b+2)}{b+1}+\frac{c(c+2)}{c+1}}\\ &= \frac{9}{\frac{(a+1)^2-1}{a+1}+\frac{(b+1)^2-1}{b+1}+\frac{(c+1)^2-1}{c+1}}\\ &= \frac{9}{(a+1)+(b+1)+(c+1) -\left(\frac1{a+1}+\frac1{b+1}+\frac1{c+1}\right)}. \end{aligned}$$

Kako je

$$(a+1)+(b+1)+(c+1)=a+b+c+3\le6$$

i

$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1} \ge\frac{9}{(a+1)+(b+1)+(c+1)} \ge\frac96=\frac32,$$

vrijedi

$$\frac{9}{(a+1)+(b+1)+(c+1) -\left(\frac1{a+1}+\frac1{b+1}+\frac1{c+1}\right)} \ge\frac{9}{6-\frac32}=2,$$

čime smo dokazali tvrdnju.

Drugo rješenje.

Vrijedi

$$\frac{x+1}{x(x+2)}=\frac12\left(\frac1x+\frac1{x+2}\right)$$

pa je

$$\begin{aligned} \frac{a+1}{a(a+2)}+\frac{b+1}{b(b+2)}+\frac{c+1}{c(c+2)} &=\frac12\left(\frac1a+\frac1b+\frac1c\right) +\frac12\left(\frac1{a+2}+\frac1{b+2}+\frac1{c+2}\right). \end{aligned}$$

Primjenom nejednakosti između aritmetičke i harmonijske sredine, dobivamo

$$\frac{a+b+c}{3}\ge\frac{3}{\frac1a+\frac1b+\frac1c}.$$

a kako je $a+b+c\le3$ slijedi

$$\frac1a+\frac1b+\frac1c\ge\frac{9}{a+b+c}\ge3.$$

Slično postupamo i s drugim dijelom. Primijenimo A–H nejednakost na brojeve $a+2$, $b+2$, $c+2$ i iskoristimo uvjet $(a+2)+(b+2)+(c+2)\le9$:

$$\frac1{a+2}+\frac1{b+2}+\frac1{c+2} \ge\frac{9}{(a+2)+(b+2)+(c+2)}\ge1.$$

Konačno imamo

$$\frac{a+1}{a(a+2)}+\frac{b+1}{b(b+2)}+\frac{c+1}{c(c+2)} \ge\frac12\cdot3+\frac12\cdot1=2.$$