Cách 5:.(Sử dụng BĐT trung gian)Trước hêt, ta c/m BĐT đại diện: $\frac{a}{\sqrt{a^2+8bc}}\geq \frac{a^{\frac{4}{3}}}{\sum_{sym}^{}a^{\frac{4}{3}} }$ $\Leftrightarrow (\sum_{sym}^{}a^{\frac{4}{3}})^2\geq a^{\frac{2}{3}}(a^2+8bc).$Thật vậy, ta có: $(\sum_{sym}^{}a^{\frac{4}{3}})^2-(a^{\frac{4}{3}})^2=(b^{\frac{4}{3}}+c^{\frac{4}{3}}).(2a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}})\geq (2b^{\frac{2}{3}}c^{\frac{4}{3}})(4a^{\frac{2}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}})=8a^{\frac{2}{3}}bc.$Suy ra:: $(\sum_{sym}^{}a^{\frac{4}{3}})^2\geq (a^{\frac{4}{3}})^2+8a^{\frac{2}{3}}bc=a^{\frac{2}{3}}(a^2+8bc).$Vậy: $\frac{a}{\sqrt{a^2+8bc}}\geq \frac{a^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}};...............$Từ đó, cộng 2 vế 3 bđt cùng chiều suy ra đpcm

Cách 5:.(Sử dụng BĐT trung gian)Trước hêt, ta c/m BĐT đại diện: $\frac{a}{\sqrt{a^2+8bc}}\geq \frac{a^{\frac{4}{3}}}{\sum_{}^{}a^{\frac{4}{3}} }$ $\Leftrightarrow (\sum_{}^{}a^{\frac{4}{3}})^2\geq a^{\frac{2}{3}}(a^2+8bc).$Thật vậy, ta có: $(\sum_{}^{}a^{\frac{4}{3}})^2-(a^{\frac{4}{3}})^2=(b^{\frac{4}{3}}+c^{\frac{4}{3}}).(2a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}})\geq (2b^{\frac{2}{3}}c^{\frac{4}{3}})(4a^{\frac{2}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}})=8a^{\frac{2}{3}}bc.$Suy ra:: $(\sum_{}^{}a^{\frac{4}{3}})^2\geq (a^{\frac{4}{3}})^2+8a^{\frac{2}{3}}bc=a^{\frac{2}{3}}(a^2+8bc).$Vậy: $\frac{a}{\sqrt{a^2+8bc}}\geq \frac{a^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}};...............$Từ đó, cộng 2 vế 3 bđt cùng chiều suy ra đpcm

Cách 4: Vận dụng BĐT BunyakoxskyTa có:$(a+b+c)^2\leq (\sum_{sym}^{}\frac{a}{\sqrt{a^2+8bc}}).(\sum_{sym}^{}a\sqrt{a^2+8bc}) =(\sum_{sym}^{}\frac{a}{\sqrt{a^2+8bc}}).(\sum_{sym}^{}\sqrt{a}\sqrt{a^3+8abc})\leq (\frac{a}{\sqrt{a^2+8bc}}).\sqrt{(a+b+c)(a^3+b^3+c^3+24abc)}\leq (\sum_{sym}^{}\frac{a}{\sqrt{a^2+8bc}}).\sqrt{(a+b+c)(a+b+c)^3}. $$\Rightarrow .................$

Cách 4: Vận dụng BĐT BunyakoxskyTa có:$(a+b+c)^2\leq (\sum_{}^{}\frac{a}{\sqrt{a^2+8bc}}).(\sum_{}^{}a\sqrt{a^2+8bc}) =(\sum_{}^{}\frac{a}{\sqrt{a^2+8bc}}).(\sum_{}^{}\sqrt{a}\sqrt{a^3+8abc})\leq (\frac{a}{\sqrt{a^2+8bc}}).\sqrt{(a+b+c)(a^3+b^3+c^3+24abc)}\leq (\sum_{}^{}\frac{a}{\sqrt{a^2+8bc}}).\sqrt{(a+b+c)(a+b+c)^3}. $$\Rightarrow .................$

Cách 3: Vận dụng BĐT Chebyshev, Cauchy, Bunyakovsky.Không giảm tính tổng quát, giả sử $a\geq b\geq c$$P\geq \sum_{sym}^{} \frac{a}{\sqrt{a^2+b^2+c^2+6bc}}\geq\frac{1}{3}.(a+b+c).(\sum_{sym}^{} \frac{1}{\sqrt{a^2+b^2+c^2+6bc}})\geq \frac{1}{3}.(a+b+c).\frac{9}{\sum_{sym}^{}\sqrt{a^2+b^2+c^2+6bc} }\geq \frac{1}{3}.(a+b+c).\frac{9}{\sqrt{3[\sum_{sym}^{} }(a^2+b^2+6bc)]}=1$$\Rightarrow ..................$

Cách 3: Vận dụng BĐT Chebyshev, Cauchy, Bunyakovsky.Không giảm tính tổng quát, giả sử $a\geq b\geq c$$P\geq \sum_{}^{} \frac{a}{\sqrt{a^2+b^2+c^2+6bc}}\geq\frac{1}{3}.(a+b+c).(\sum_{}^{} \frac{1}{\sqrt{a^2+b^2+c^2+6bc}})\geq \frac{1}{3}.(a+b+c).\frac{9}{\sum_{}^{}\sqrt{a^2+b^2+c^2+6bc} }\geq \frac{1}{3}.(a+b+c).\frac{9}{\sqrt{3[\sum_{}^{} }(a^2+b^2+6bc)]}=1$$\Rightarrow ..................$