=> P = $ \frac{\frac{1}{a^{2}}}{\frac{1}{b} + \frac{1}{c}}$ + $ \frac{\frac{1}{b^{2}}}{\frac{1}{a}+ \frac{1}{c} }$ + $ \frac{\frac{1}{c^{2}}}{\frac{1}{a} + \frac{1}{b}}$ Đặt : \begin{cases}x= \frac{1}{a} \\ y=\frac{1}{b} => xyz=1 => P = \frac{x^{2}}{y+z} + \frac{y^{2}}{z+x} + \frac{z^{2}}{x+y}\\z= \frac{1}{c} \end{cases}Áp dụng BĐT Cô-si : $\frac{x^{2}}{y+z}$ + $\frac{y+z}{4}$ $\geq$ x=> Tương tự : $\frac{y^{2}}{z+x}$ + $ \frac{z+x}{4}$ $\geq $ y ; $ \frac{z^{2}}{x+y}$ + $ \frac{x+y}{4}$ $\geq $ zTừ đó => P = $\frac{x^{2}}{y+z}$ + $\frac{y^{2}}{z+x}$ +$\frac{z^{2}}{x+y}$ $\geq $ $\frac{x+y+z}{2}$ $\geq $ $\frac{3\sqrt[3]{xyz} }{2}$ = $\frac{3}{2}$Dâu "=" xảy ra <=> x=y=z <=> a=b=c=1 Vậy Min P = $\frac{3}{2}$ khi a=b=c=1
=> P = $ \frac{\frac{1}{a^{2}}}{\frac{1}{b} + \frac{1}{c}}$ + $ \frac{\frac{1}{b^{2}}}{\frac{1}{a}+ \frac{1}{c} }$ + $ \frac{\frac{1}{c^{2}}}{\frac{1}{a} + \frac{1}{b}}$ Đặt : \begin{cases}x= \frac{1}{a} \\ y=\frac{1}{b} => xyz=1 => P = \frac{x^{2}}{y+z} + \frac{y^{2}}{z+x} + \frac{z^{2}}{x+y}\\z= \frac{1}{c} \end{cases}Áp dụng BĐT Cô-si : $\frac{x^{2}}{y+z}$ + $\frac{y+z}{4}$ $\geq$ x=> Tương tự : $\frac{y^{2}}{z+x}$ + $ \frac{z+x}{4}$ $\geq $ y ; $ \frac{z^{2}}{x+y}$ + $ \frac{x+y}{4}$ $\geq $ zTừ đó => P = $\frac{x^{2}}{y+z}$ + $\frac{y^{2}}{z+x}$ +$\frac{z^{2}}{x+y}$ $\geq $ $\frac{x+y+z}{2}$ $\geq $ $\frac{3\sqrt[3]{xyz} }{2}$ = $\frac{3}{2}$Dâu "=" xảy ra <=> x=y=z <=> a=b=c=1 Vậy Min P = $\frac{3}{2} khi a=b=c=1
=> P = $ \frac{\frac{1}{a^{2}}}{\frac{1}{b} + \frac{1}{c}}$ + $ \frac{\frac{1}{b^{2}}}{\frac{1}{a}+ \frac{1}{c} }$ + $ \frac{\frac{1}{c^{2}}}{\frac{1}{a} + \frac{1}{b}}$ Đặt : \begin{cases}x= \frac{1}{a} \\ y=\frac{1}{b} => xyz=1 => P = \frac{x^{2}}{y+z} + \frac{y^{2}}{z+x} + \frac{z^{2}}{x+y}\\z= \frac{1}{c} \end{cases}Áp dụng BĐT Cô-si : $\frac{x^{2}}{y+z}$ + $\frac{y+z}{4}$ $\geq$ x=> Tương tự : $\frac{y^{2}}{z+x}$ + $ \frac{z+x}{4}$ $\geq $ y ; $ \frac{z^{2}}{x+y}$ + $ \frac{x+y}{4}$ $\geq $ zTừ đó => P = $\frac{x^{2}}{y+z}$ + $\frac{y^{2}}{z+x}$ +$\frac{z^{2}}{x+y}$ $\geq $ $\frac{x+y+z}{2}$ $\geq $ $\frac{3\sqrt[3]{xyz} }{2}$ = $\frac{3}{2}$Dâu "=" xảy ra <=> x=y=z <=> a=b=c=1 Vậy Min P = $\frac{3}{2}
$ khi a=b=c=1