help me!giup toi lam bai BDT nay cai

K

khanh_ndd

bài này tui vừa post bên box bđt 10. OK!!
Ta luôn có [TEX]\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{3}{abc}[/TEX]
Mặt # [TEX](ab+bc+ca)^2\geq 3abc(a+b+c) \Rightarrow \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{3}{abc}\geq \frac{27}{(ab+bc+ca)^2}[/TEX](do [TEX]a+b+c=3[/TEX]
nên ta cần cm [TEX]\frac{27}{(ab+bc+ca)^2}\geq a^2+b^2+c^2[/TEX] hay [TEX](ab+bc+ca)^2.(a^2+b^2+c^2)\leq 27[/TEX]
Áp dụng bđt [TEX]xyz\leq (\frac{x+y+z}{3})^3[/TEX] ta có
[TEX](ab+bc+ca)^2.(a^2+b^2+c^2)\leq (\frac{2(ab+bc+ca)+a^2+b^2+c^2}{3})^3=(\frac{(a+b+c)^2}{3})^3=27\Rightarrow [/TEX]đpcm
Dấu = khi a=b=c=1
 
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