View attachment 134986
The†FireᴥSwordᵛᶥᶯᶣ†††♥♥♥♪ Giúp e với
[tex]P=\frac{a+b+c}{\sqrt{abc}}+2\sqrt{abc}.(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\\\\ =\frac{a+b+c}{\sqrt{abc}}+\frac{2\sqrt{bc}}{\sqrt{a}}+\frac{2\sqrt{ca}}{\sqrt{b}}+\frac{2\sqrt{ab}}{\sqrt{c}}\\\\ =\frac{a+b+c}{\sqrt{abc}}+2.\frac{bc+ca+ab}{\sqrt{abc}}\\\\ +, (bc+ca+ab)^2\geq 3abc.(a+b+c)=9abc\\\\ => bc+ca+ab\geq 3\sqrt{abc}\\\\ +,3=a+b+c\geq 3\sqrt[3]{abc}=> abc\leq 1 => \sqrt{abc}\leq 1\\\\ => P\geq \frac{3}{1}+2.\frac{3\sqrt{abc}}{\sqrt{abc}}=3+6=9[/tex]
dấu "=" <=> a=b=c=1
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