mình giúp bạn nhé
Ta có
[TEX]\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\geq \frac{9}{ab+bc+ca}[/TEX]
nên [TEX]\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\geq \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} +\frac{9}{ab+bc+ca} [/TEX]
Ta có [TEX]\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} +\frac{9}{ab+bc+ca} =\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} +\frac{1}{ab+bc+ca}+ \frac{1}{ab+bc+ca}+\frac{7}{ab+bc+ca} [/TEX]
mà
[TEX]\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} +\frac{1}{ab+bc+ca}+ \frac{1}{ab+bc+ca}\geq\frac{9}{(a+b+c)^2}=9 (1)[/TEX]
[TEX]\frac{7}{ab+bc+ca}\geq \frac{21}{(a+b+c)^2}=21 (2)[/TEX]
Từ (1) và (2) suy ra
[TEX]P\geq 30[/TEX]
dấu bằng xảy ra khi [TEX]a=b=c = \frac{1}{3}[/TEX]