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