ad BĐT C-Sta có $(a^{2}+b+c)(1+b+c)\geq (a+b+c)^{2}$$\Rightarrow \sqrt{\frac{a^{2}}{a^{2}+b+c}}\leq \frac{a\sqrt{b+c+1}}{a+b+c}=\frac{a\sqrt{3(b+c+1)}}{3\sqrt{3}}\leq \frac{a+(b+c+1+3)}{2(a+b+c)\sqrt{3}}$TT$\Rightarrow VT\leq \frac{4(a+b+c)+2(ab+bc+ca)}{2(a+b+c)\sqrt{3}}=\frac{2\sqrt{3}}{3}+\frac{1}{\sqrt{3}} \frac{ab+bc+ca}{a+b+c}=\frac{2\sqrt{3}}{3}+\frac{1}{3}\sqrt{ab+bc+ca}\leq \sqrt{3}$Dấu "="$\Leftrightarrow a=b=c=1$

ad BĐT C-Sta có $(a^{2}+b+c)(1+b+c)\geq (a+b+c)^{2}$$\Rightarrow \sqrt{\frac{a^{2}}{a^{2}+b+c}}\leq \frac{a\sqrt{b+c+1}}{a+b+c}=\frac{a\sqrt{3(b+c+1)}}{(a+b+c)\sqrt{3}}\leq \frac{a+(b+c+1+3)}{2(a+b+c)\sqrt{3}}$TT$\Rightarrow VT\leq \frac{4(a+b+c)+2(ab+bc+ca)}{2(a+b+c)\sqrt{3}}=\frac{2\sqrt{3}}{3}+\frac{1}{\sqrt{3}} \frac{ab+bc+ca}{a+b+c}=\frac{2\sqrt{3}}{3}+\frac{1}{3}\sqrt{ab+bc+ca}\leq \sqrt{3}$Dấu "="$\Leftrightarrow a=b=c=1$

ad BĐT C-Sta có $(a^{2}+b+c)(1+b+c)\geq (a+b+c)^{2}$$\Rightarrow \sqrt{\frac{a^{2}}{a^{2}+b+c}}\leq \frac{a\sqrt{b+c+1}}{3}=\frac{a\sqrt{3(b+c+1)}}{3\sqrt{3}}\leq \frac{a+(b+c+1+3)}{6\sqrt{3}}$TT$\Rightarrow VT\leq \frac{4(a+b+c)+2(ab+bc+ca)}{6\sqrt{3}}\leq \frac{4.3+2.3}{6\sqrt{3}}=\sqrt{3}$Dấu "="$\Leftrightarrow a=b=c=1$

ad BĐT C-Sta có $(a^{2}+b+c)(1+b+c)\geq (a+b+c)^{2}$$\Rightarrow \sqrt{\frac{a^{2}}{a^{2}+b+c}}\leq \frac{a\sqrt{b+c+1}}{a+b+c}=\frac{a\sqrt{3(b+c+1)}}{3\sqrt{3}}\leq \frac{a+(b+c+1+3)}{2(a+b+c)\sqrt{3}}$TT$\Rightarrow VT\leq \frac{4(a+b+c)+2(ab+bc+ca)}{2(a+b+c)\sqrt{3}}=\frac{2\sqrt{3}}{3}+\frac{1}{\sqrt{3}} \frac{ab+bc+ca}{a+b+c}=\frac{2\sqrt{3}}{3}+\frac{1}{3}\sqrt{ab+bc+ca}\leq \sqrt{3}$Dấu "="$\Leftrightarrow a=b=c=1$