Câu 1: $\sum_{cyc}^{a,b,c}\sqrt{a^2+abc}=\sum_{cyc}^{a,b,c}\sqrt{a(a(a+b+c)+bc)}=\sum_{}^{} \sqrt{a(a+b)(a+c)} $
Mà $\sum_{}^{}\sqrt{a(a+b)(a+c)}\le\sum_{}^{} \frac{\sqrt{a}}{2}(2a+b+c)=\sum_{}^{} \frac{\sqrt{a}}{2}(a+1) $
Ta có: $\frac{\sqrt{a}}{2}(a+1)+\sqrt{abc}\le\frac{\sqrt{a}}{2}(a+1+b+c)=\sqrt{a}$
Khi đó $M\le\sum_{}^{} \sqrt{a}+6\sqrt{abc}$
Lại có $\sum \sqrt{a}=\sum \sqrt{3}.\sqrt{\frac{a}{3}}\le\frac{\sqrt{3}}{2}(a+b+c+1)=\sqrt{3}$
Vậy $M\le\sqrt{3}+6\sqrt{abc}\le\sqrt{3}+6\sqrt{\frac{(a+b+c)^3}{27}}=\frac{5\sqrt{3}}{3}$
Dấu bằng có khi $a=b=c=\frac{1}{3}$