Školjka

\begin{align*} &\frac{1}{3} \left(\frac{a^3-b^3}{a^2-b^2}+\frac{b^3-c^3}{b^2-c^2}+\frac{c^3-a^3}{c^2-a^2}\right)\\ &=\frac{1}{3}\left(\sum_{cyc}\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a+b)}\right)\\ &=\frac{1}{3}\left(\sum_{cyc}\frac{a^2+ab+b^2}{a+b}\right)\\ &=\frac{1}{3}\left(\sum_{cyc}\frac{a^2}{a+b}+b\right)\\ &\stackrel{\text{AG}}{\le}\frac{1}{3}\left(\sum_{cyc}\frac{a^2}{a+b}+2\frac{a^2}{a+b}\right)\\ &=\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a} \end{align*} Nejednakost slijedi iz $AG$-a: $$\sum_{cyc}\left(\frac{a^2}{a+b}+\frac{a+b}{4}\right)\ge \sum_{cyc}a$$ $$\Longleftrightarrow 2\left(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\right)\ge a+b+c $$ Ako se jednakost postiže, tada imamo $3a^2=2ab+b^2$\\ $3b^2=2bc+c^2$\\ $3c^2=2ac+a^2$\\ Zbrajajući dobivamo $a^2+b^2+c^2=ab+bc+ca$, što vrijedi akko $a=b=c.\Longrightarrow\Longleftarrow$. Dakle, $$\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a} >\frac{1}{3} \left(\frac{a^3-b^3}{a^2-b^2}+\frac{b^3-c^3}{b^2-c^2}+\frac{c^3-a^3}{c^2-a^2}\right)$$