Theo BĐT AM-GM cho 2 số thực dương ta có$\sqrt{bc(b^{2}+c^{2})}$= $\frac{1}{\sqrt{2}}$. $\sqrt{2bc(b^{2}+c^{2})}$ $\leq$ $\frac{2bc}{2\sqrt{2}}$+$\frac{b^{2}}{2\sqrt{2}}$+$\frac{c^{2}}{2\sqrt{2}}$=$\frac{(b+c)^{2}}{2\sqrt{2}}$$\Rightarrow \frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}} \geq \frac{2\sqrt{2}a(b+c)}{(b+c)^{2}} = \frac{2\sqrt{2}a}{b+c} $tương tự:$\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}}\geq\frac{2\sqrt{2}b}{c+a}$ $\frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}} \geq \frac{2\sqrt{2}c}{a+b}$cộng cả 3 vế lại ta có:$\frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}}+\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}}\geq \frac{2\sqrt{2}b}{c+a} +\frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}}\geq\frac{2\sqrt{2}c}{a+b}\geq 2\sqrt{2}(\frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b})\geq2\sqrt{2}.\frac{3}{2}=3\sqrt{2}$

Theo BĐT AM-GM cho 2 số thực dương ta có$\sqrt{bc(b^{2}+c^{2})}=\frac{1}{\sqrt{2}}.\sqrt{2bc(b^{2}+c^{2})}$ $\leq \frac{2bc}{2\sqrt{2}}+\frac{b^{2}}{2\sqrt{2}}+\frac{c^{2}}{2\sqrt{2}}$=$\frac{(b+c)^{2}}{2\sqrt{2}}$$\Rightarrow \frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}} \geq \frac{2\sqrt{2}a(b+c)}{(b+c)^{2}} = \frac{2\sqrt{2}a}{b+c} $tương tự:$\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}}\geq\frac{2\sqrt{2}b}{c+a}$ $\frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}} \geq \frac{2\sqrt{2}c}{a+b}$cộng cả 3 vế lại ta có:$\frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}}+\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}}\geq \frac{2\sqrt{2}b}{c+a} +\frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}}\geq\frac{2\sqrt{2}c}{a+b}\geq 2\sqrt{2}(\frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b})\geq2\sqrt{2}.\frac{3}{2}=3\sqrt{2}$

Theo BĐT AM-GM cho 2 số thực dương ta có$\sqrt{bc(b^{2}+c^{2})}$=$\frac{1}{\sqrt{2}}$. $\sqrt{2bc(b^{2}+c^{2})}$ $\leq$ $\frac{2bc}{2\sqrt{2}}$+$\frac{b^{2}}{2\sqrt{2}}$+$\frac{c^{2}}{2\sqrt{2}}$=$\frac{(b+c)^{2}}{2\sqrt{2}}$$\Rightarrow \frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}} \geq \frac{2\sqrt{2}a(b+c)}{(b+c)^{2}} = \frac{2\sqrt{2}a}{b+c} $tương tự:$\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}}\geq\frac{2\sqrt{2}b}{c+a}$ $\frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}} \geq \frac{2\sqrt{2}c}{a+b}$cộng cả 3 vế lại ta có:$\frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}}+\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}}\geq \frac{2\sqrt{2}b}{c+a} +\frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}}\geq\frac{2\sqrt{2}c}{a+b}\geq 2\sqrt{2}(\frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b})\geq2\sqrt{2}.\frac{3}{2}=3\sqrt{2}$

Theo BĐT AM-GM cho 2 số thực dương ta có$\sqrt{bc(b^{2}+c^{2})}$= $\frac{1}{\sqrt{2}}$. $\sqrt{2bc(b^{2}+c^{2})}$ $\leq$ $\frac{2bc}{2\sqrt{2}}$+$\frac{b^{2}}{2\sqrt{2}}$+$\frac{c^{2}}{2\sqrt{2}}$=$\frac{(b+c)^{2}}{2\sqrt{2}}$$\Rightarrow \frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}} \geq \frac{2\sqrt{2}a(b+c)}{(b+c)^{2}} = \frac{2\sqrt{2}a}{b+c} $tương tự:$\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}}\geq\frac{2\sqrt{2}b}{c+a}$ $\frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}} \geq \frac{2\sqrt{2}c}{a+b}$cộng cả 3 vế lại ta có:$\frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}}+\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}}\geq \frac{2\sqrt{2}b}{c+a} +\frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}}\geq\frac{2\sqrt{2}c}{a+b}\geq 2\sqrt{2}(\frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b})\geq2\sqrt{2}.\frac{3}{2}=3\sqrt{2}$

Theo BĐT AM-GM cho 2 số thực dương ta có$\sqrt{bc(b^{2}+c^{2})} = \frac{1}{\sqrt{2}}.\sqrt{2bc(b^{2}+c^{2})} \leq \frac{2bc+b^{2}+c^{2}}{2\sqrt{2}} =\frac{(b+c)^{2}}{2\sqrt{2}}$$\Rightarrow \frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}} \geq \frac{2\sqrt{2}a(b+c)}{(b+c)^{2}} = \frac{2\sqrt{2}a}{b+c} $tương tự:$\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}}\geq\frac{2\sqrt{2}b}{c+a}$ $\frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}} \geq \frac{2\sqrt{2}c}{a+b}$cộng cả 3 vế lại ta có:$\frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}}+\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}}\geq \frac{2\sqrt{2}b}{c+a} +\frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}}\geq\frac{2\sqrt{2}c}{a+b}\geq 2\sqrt{2}(\frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b})\geq2\sqrt{2}.\frac{3}{2}=3\sqrt{2}$

Theo BĐT AM-GM cho 2 số thực dương ta có$\sqrt{bc(b^{2}+c^{2})}$=$\frac{1}{\sqrt{2}}$. $\sqrt{2bc(b^{2}+c^{2})}$ $\leq$ $\frac{2bc}{2\sqrt{2}}$+$\frac{b^{2}}{2\sqrt{2}}$+$\frac{c^{2}}{2\sqrt{2}}$=$\frac{(b+c)^{2}}{2\sqrt{2}}$$\Rightarrow \frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}} \geq \frac{2\sqrt{2}a(b+c)}{(b+c)^{2}} = \frac{2\sqrt{2}a}{b+c} $tương tự:$\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}}\geq\frac{2\sqrt{2}b}{c+a}$ $\frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}} \geq \frac{2\sqrt{2}c}{a+b}$cộng cả 3 vế lại ta có:$\frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}}+\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}}\geq \frac{2\sqrt{2}b}{c+a} +\frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}}\geq\frac{2\sqrt{2}c}{a+b}\geq 2\sqrt{2}(\frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b})\geq2\sqrt{2}.\frac{3}{2}=3\sqrt{2}$

Theo BĐT AM-GM cho 2 số thực dương ta có$ \sqrt{bc(b^{2}+c^{2})} = \frac{1}{\sqrt{2}}. \sqrt{2bc(b^{2}+c^{2})} \leq \frac{2bc+b^{2}+c^{2}}{2\sqrt{2}} =\frac{(b+c)^{2}}{2\sqrt{2}}\Rightarrow \frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}} \geq \frac{2\sqrt{2}a(b+c)}{(b+c)^{2}} = \frac{2\sqrt{2}a}{b+c} $tương tự:$\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}} \geq \frac{2\sqrt{2}b}{c+a} \frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}} \geq \frac{2\sqrt{2}c}{a+b}$cộng cả 3 vế lại ta có:$\frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}} + \frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}} \geq \frac{2\sqrt{2}b}{c+a} + \frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}} \geq \frac{2\sqrt{2}c}{a+b} \geq 2\sqrt{2}(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}) \geq 2\sqrt{2}.\frac{3}{2} = 3\sqrt{2} $

Theo BĐT AM-GM cho 2 số thực dương ta có$\sqrt{bc(b^{2}+c^{2})} = \frac{1}{\sqrt{2}}.\sqrt{2bc(b^{2}+c^{2})} \leq \frac{2bc+b^{2}+c^{2}}{2\sqrt{2}} =\frac{(b+c)^{2}}{2\sqrt{2}}$$\Rightarrow \frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}} \geq \frac{2\sqrt{2}a(b+c)}{(b+c)^{2}} = \frac{2\sqrt{2}a}{b+c} $tương tự:$\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}}\geq\frac{2\sqrt{2}b}{c+a}$ $\frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}} \geq \frac{2\sqrt{2}c}{a+b}$cộng cả 3 vế lại ta có:$\frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}}+\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}}\geq \frac{2\sqrt{2}b}{c+a} +\frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}}\geq\frac{2\sqrt{2}c}{a+b}\geq 2\sqrt{2}(\frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b})\geq2\sqrt{2}.\frac{3}{2}=3\sqrt{2}$

Theo BĐT AM-GM cho 2 số thực dương ta có$ \sqrt{bc(b^{2}+c^{2})} = \frac{1}{\sqrt{2}}. \sqrt{2bc(b^{2}+c^{2})} \leq \frac{2bc+b^{2}+c^{2}}{2\sqrt{2}} =\frac{(b+c)^{2}}{2\sqrt{2}}\Rightarrow \frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}} \geq \frac{2\sqrt{2}a(b+c)}{(b+c)^{2}} = \frac{2\sqrt{2}a}{b+c} $tương tự:$\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}} \geq \frac{2\sqrt{2}b}{c+a} \frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}} \geq \frac{2\sqrt{2}c}{a+b}$cộng cả 3 vế lại ta có:$\frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}} + \frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}} \geq \frac{2\sqrt{2}b}{c+a} + \frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}} \geq \frac{2\sqrt{2}c}{a+b} \geq 2\sqrt{2}(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}) \geq 2\sqrt{2}.\frac{3}{2} = 3\sqrt{2} $

Theo BĐT AM-GM cho 2 số thực dương ta có$ \sqrt{bc(b^{2}+c^{2})} = \frac{1}{\sqrt{2}}. \sqrt{2bc(b^{2}+c^{2})} \leq \frac{2bc+b^{2}+c^{2}}{2\sqrt{2}} =\frac{(b+c)^{2}}{2\sqrt{2}}\Rightarrow \frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}} \geq \frac{2\sqrt{2}a(b+c)}{(b+c)^{2}} = \frac{2\sqrt{2}a}{b+c} $tương tự:$\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}} \geq \frac{2\sqrt{2}b}{c+a} \frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}} \geq \frac{2\sqrt{2}c}{a+b}$cộng cả 3 vế lại ta có:$\frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}} + \frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}} \geq \frac{2\sqrt{2}b}{c+a} + \frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}} \geq \frac{2\sqrt{2}c}{a+b} \geq 2\sqrt{2}(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}) \geq 2\sqrt{2}.\frac{3}{2} = 3\sqrt{2} $

Theo BĐT AM-GM cho 2 số thực dương ta có$ \sqrt{bc(b^{2}+c^{2})} = \frac{1}{\sqrt{2}}. \sqrt{2bc(b^{2}+c^{2})} \leq \frac{2bc+b^{2}+c^{2}}{2\sqrt{2}} =\frac{(b+c)^{2}}{2\sqrt{2}}\Rightarrow \frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}} \geq \frac{2\sqrt{2}a(b+c)}{(b+c)^{2}} = \frac{2\sqrt{2}a}{b+c} tương tự:\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}} \geq \frac{2\sqrt{2}b}{c+a} \frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}} \geq \frac{2\sqrt{2}c}{a+b}cộng cả 3 vế lại ta có:\frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}} + \frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}} \geq \frac{2\sqrt{2}b}{c+a} + \frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}} \geq \frac{2\sqrt{2}c}{a+b} \geq 2\sqrt{2}(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}) \geq 2\sqrt{2}.\frac{3}{2} = 3\sqrt{2} $

Theo BĐT AM-GM cho 2 số thực dương ta có$ \sqrt{bc(b^{2}+c^{2})} = \frac{1}{\sqrt{2}}. \sqrt{2bc(b^{2}+c^{2})} \leq \frac{2bc+b^{2}+c^{2}}{2\sqrt{2}} =\frac{(b+c)^{2}}{2\sqrt{2}}\Rightarrow \frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}} \geq \frac{2\sqrt{2}a(b+c)}{(b+c)^{2}} = \frac{2\sqrt{2}a}{b+c} $tương tự:$\frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}} \geq \frac{2\sqrt{2}b}{c+a} \frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}} \geq \frac{2\sqrt{2}c}{a+b}$cộng cả 3 vế lại ta có:$\frac{a(b+c)}{\sqrt{bc(b^{2}+c^{2})}} + \frac{b(c+a)}{\sqrt{ca(c^{2}+a^{2})}} \geq \frac{2\sqrt{2}b}{c+a} + \frac{c(a+b)}{\sqrt{ab(a^{2}+b^{2})}} \geq \frac{2\sqrt{2}c}{a+b} \geq 2\sqrt{2}(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}) \geq 2\sqrt{2}.\frac{3}{2} = 3\sqrt{2} $