Phải chuẩn hóa $a+ b+ c= 3$, thế nên ta được:
$$\sum\limits_{cyc}\frac{\left ( b+ c- a \right )^{2}}{2\,a^{2}+ \left ( b+ c \right )^{2}}\geqq \frac{3\left ( \sum\limits_{cyc}a^{2} \right )}{2\left ( a+ b+ c \right )^{2}}$$
$$\Leftrightarrow \sum\limits_{cyc} \frac{\left ( 3- 2\,a \right )^{2}}{2\,a^{2}+ \left ( 3- a \right )^{2}} \geqq \frac{\sum\limits_{cyc}a^{2}}{6}$$
$$\Leftrightarrow \sum\limits_{cyc} \frac{\left ( 1- a \right )\left ( a+ 3 \right )\left ( a^{2}- 4\,a+ 6 \right )}{6\left ( a^{2}- 2\,a+ 3 \right )}\geqq 0$$
$$\Leftrightarrow \sum\limits_{cyc} \frac{\left ( 1- a \right )\left ( a+ 3 \right )\left ( a^{2}- 4\,a+ 6 \right )}{6\left ( a^{2}- 2\,a+ 3 \right )}+ \sum\limits_{cyc}\left ( a- 1 \right )\geqq 0$$
$$\Leftrightarrow \sum\limits_{cyc} \frac{\left ( 6- a \right )a\left ( a- 1 \right )^{2}}{6\left ( a^{2}- 2\,a+ 3 \right )}\geqq 0$$