đổi biến cho dễ nhìn thay$\sqrt{a}=a,\sqrt{b}=b,\sqrt{c}=c$
$\Rightarrow a+b+c=1$
cần chứng minh $A=\sum \frac{a^{4}+b^{2}c^{2}}{a^{2}\sqrt{2\left ( b^{2}+c^{2} \right )}}\geq 1$
$A=\sum \frac{a^{2}}{\sqrt{2\left ( b^{2}+c^{2} \right )}}+\sum \frac{\frac{b^{2}c^{2}}{a^{2}}}{\sqrt{2\left ( b^{2}+c^{2} \right )}}\geq 1$
xét $B=\sum \frac{a^{2}}{\sqrt{\left ( b^{2}+c^{2} \right )}}$
đặt $\sqrt{a^{2}+b^{2}}=x,\sqrt{b^{2}+c^{2}}=y,\sqrt{c^{2}+a^{2}}=z$
$B=\sum \frac{y^{2}+z^{2}-x^{2}}{2x}= \sum \frac{y^{2}}{2x}+\sum \frac{z^{2}}{x}-x-y-z\geq \frac{x+y+z}{2}=\frac{\sum \sqrt{a^{2}+b^{2}}}{2}\geq \sum \frac{a+b}{2\sqrt{2}}=\frac{a+b+c}{\sqrt{2}}$
$\Rightarrow \frac{a^{2}}{\sqrt{2\left ( b^{2}+c^{2} \right )}}\geq \frac{a+b+c}{2}=\frac{1}{2}$ $\left ( 1 \right )$
xét $C= \frac{\frac{b^{2}c^{2}}{a^{2}}}{\sqrt{2\left ( b^{2}+c^{2} \right )}}\geq \frac{\left ( \sum \frac{bc}{a} \right )^{2}}{\sum \sqrt{2\left ( b^{2}+c^{2} \right )}}$
ta chứng minh $\left ( \sum \frac{ab}{c} \right )\geq 3\sum a^{2}$
$\Leftrightarrow \sum \frac{a^{2}b^{2}}{c^{2}}+2\sum a^{2}\geq 3\sum a^{2}$ (đúng)
$\Rightarrow C\geq \frac{3\left ( a^{2}+b^{2}+c^{2} \right )}{\sqrt{2\left ( b^{2}+c^{2} \right )}}$
$\Rightarrow C\geq \frac{\sum a^{2}}{\sqrt{\frac{2}{9}\left ( b^{2}+c^{2} \right )}}\geq \frac{2\sum a^{2}}{2\sum a^{2}+\frac{2}{3}}=\frac{2}{2+\frac{2}{3\sum a^{2}}}\geq \frac{2}{2+\frac{2}{\left ( a+b+c \right )^{2}}}=\frac{1}{2}$ $\left ( 2 \right )$
từ $\left ( 1 \right ),\left ( 2 \right )$ suy ra dpcm