giúp em bài nầy nha mấy bác!!!!!!!

L

lethiquynhhien

[TEX]VT=\frac{1}{(a+c)+2(b+c)}+\frac{1}{(b+c)+2(a+b)}[/TEX][TEX]+\frac{1}{(c+a)+2(a+c)}[/TEX]
Áp dụng BDT phụ: [TEX]\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\frac{9}{a+b+c}[/TEX]ta có:
[TEX]9VT\leq\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{b+c}[/TEX][TEX]+\frac{1}{b+c}+\frac{1}{a+b}+\frac{1}{a+b}+\frac{1}{b+a}+\frac{1}{a+c}+\frac{1}{a+c}=3(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})[/TEX] (1)
áp dụng BDT phụ: [TEX]\frac{1}{a}+\frac{1}{b}\geq\frac{4}{a+b}[/TEX]ta có:
[TEX]3(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})\leq3(\frac{1}{4})(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}[/TEX][TEX]+\frac{1}{c}+\frac{1}{a})=\frac{3}{2}[/TEX] (2)
từ (1) và (2) ta có VT[TEX]\leq\frac{3}{18}<\frac{3}{16}[/TEX]
 
B

botvit

[TEX]VT=\frac{1}{(a+c)+2(b+c)}+\frac{1}{(b+c)+2(a+b)}[/TEX][TEX]+\frac{1}{(c+a)+2(a+c)}[/TEX]
Áp dụng BDT phụ: [TEX]\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\frac{9}{a+b+c}[/TEX]ta có:
[TEX]9VT\leq\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{b+c}[/TEX][TEX]+\frac{1}{b+c}+\frac{1}{a+b}+\frac{1}{a+b}+\frac{1}{b+a}+\frac{1}{a+c}+\frac{1}{a+c}=3(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})[/TEX] (1)
áp dụng BDT phụ: [TEX]\frac{1}{a}+\frac{1}{b}\geq\frac{4}{a+b}[/TEX]ta có:
[TEX]3(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})\leq3(\frac{1}{4})(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a})=\frac{3}{2}[/TEX] (2)
từ (1) và (2) ta có VT[TEX]\leq\frac{3}{18}<\frac{3}{16}[/TEX]
chỗ đó có phải là
thế [TEX]\frac{1}{a}+\frac{1}{b}\geq\frac{4}{a+b}[/TEX]ta có:
[TEX]3(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})[/tex]
[tex]\leq3(\frac{1}{4})(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a})=\frac{3}{2}[/TEX] (2)[/tex]
 
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T

trung0123

giúp em bài này nha mấy bác!!!!!!!!!
Cho a ; b; c>0 và abc= ab+bc+ac CMR [TEX]\frac{1}{a + 2b+3c}[/TEX] +[TEX]\frac{1}{2a+3b+c}[/TEX]+[TEX]\frac{1}{3a+b+2c}[/TEX]<[TEX]\frac{3}{16}[/TEX]

áp dụng liên tiếp 2 lần BDT [TEX]\blue \frac{1}{a+b}\leq \frac{1}{4}(\frac{1}{a}+\frac{1}{b})[/TEX]
[TEX]\blue\frac{1}{a+2b+3c}=[/TEX][TEX]\blue\frac{1}{a+c+2(b+c)}\leq[/TEX] [TEX]\blue\frac{1}{4}(\frac{1}{4}(\frac{1}{a}[/TEX][TEX]\blue+\frac{1}{c})+\frac{1}{4}(\frac{1}{2b}+\frac{1}{2c}))[/TEX]
cứ như vậy cộng lại
[TEX]\red\Rightarrow dpcm[/TEX]
 
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