By C-S, \[ \left(\sum a\sqrt{\frac{b^2+c^2}2}\right)^2\le3\sum a^2\cdot\frac{b^2+c^2}2=3\sum a^2b^2. \]By homogenization, it suffices to prove \[3\sum a^2b^2\left(\sum a\right)^2\le\left(2\sum a\sum ab-9abc\right)^2.\]This is true because equivalent to \begin{align*}6abc\sum a(a-b)(a-c)+4\sum ac(b-a)^2(b-c)^2\\+\frac13\sum(b-c)^2\left(18a^2bc+(c-a)^2(a-b)^2\right)\ge0.\end{align*} For this part: youthdoo wrote: By homogenization, it suffices to prove \[3\sum a^2b^2\left(\sum a\right)^2\le\left(2\sum a\sum ab-9abc\right)^2.\] There's another way: it is equivalent to \[2\sum ab(a-c)^2(b-c)^2+6\sum a^4bc+\sum(a^4b^2+a^2b^4)-4\sum(a^3b^2c+a^3bc^2),\]which is true by Murihead.

I think the solution is very direct...

Do you have solution that is easier

Inequality is equivalent to: $4(ab+bc+ca)-6abc\ge \sum_{cyc}^{}2a\sqrt{\frac{b^2+c^2}{2}}$ $RHS=\sum_{cyc}^{}a*2\sqrt{\frac{b+c}{2}*\frac{b^2+c^2}{b+c}} \le \sum_{cyc}^{}a(\frac{b+c}{2}+\frac{b^2+c^2}{b+c})=ab+bc+ca+\sum_{cyc}^{}a(b+c-\frac{2bc}{b+c})=3(ab+bc+ca)-\sum_{cyc}^{}\frac{2abc}{b+c}$ We now need to prove: $ab+bc+ca+\sum_{cyc}^{}\frac{2abc}{b+c}\ge 6abc$ From $a+b+c=3$ we get $abc\le 1$ Therefore $ab+bc+ca\ge 3\sqrt [3]{a^2b^2c^2}\ge 3abc$ And by C-S: $2abc(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b})\ge 2abc\frac{9}{2a+2b+2c}=3abc$ Summing up following inequalities finishes the proof. Equality when $a=b=c=1$