3. Zadatak

Sređivanjem izraza na lijevoj strani dane nejednakosti dobivamo

$$\begin{aligned} \left(1+\frac ca\right)\left(1+\frac cb\right) &=\frac{(a+c)(b+c)}{ab}\\ &=\frac{c^2+ab+c(a+b)}{ab}\\ &=\frac{c^2}{ab}+1+\frac{c(a+b)}{ab}. \end{aligned}$$

Kako je trokut pravokutan, vrijedi $a^2+b^2=c^2$

$$\frac{a^2+b^2}{ab}+1+\frac{\sqrt{a^2+b^2}(a+b)}{ab}.$$

Sada koristimo $a^2+b^2\ge2ab$ i $a+b\ge2\sqrt{ab}$.

$$\begin{aligned} &\ge\frac{2ab}{ab}+1+\frac{\sqrt{2ab}\cdot2\sqrt{ab}}{ab}\\ &=2+1+2\sqrt2=3+2\sqrt2. \end{aligned}$$