[Toán 8] Bđt

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tomorrowhero

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tinhbanonlinevp447

tomorrowhero said:
1.
Cho abc=1 và [TEX]a^3 >36 [/TEX]
CMR
[TEX]\frac{a^2}{3} + b^2 + c^2> ab+bc+ac[/TEX]

2. Cmr:
[TEX]a^2 + b^2 +1 \geq ab+ a+b[/TEX]

3. Cho
a,b,c,d>0
[TEX]\frac{1}{1+a} + \frac{1}{1+b} +\frac{1}{1+c}+\frac{1}{1+d} \geq 3[/TEX]

CMR [TEX] abcd \leq \frac{1}{81}[/TEX]
Bấm để xem đầy đủ nội dung ...
1,
[TEX]abc=1 \Rightarrow bc=\frac{1}{a} [/TEX]
[TEX]a^3>36 [/TEX]nên a>0
[TEX]\frac{a^2}{3} + b^2 + c^2> ab+bc+ac[/TEX]
[TEX](\frac{a}{2}-b-c)^2+\frac{a^2}{12}+3bc > 0[/TEX]
[TEX]\Rightarrow (\frac{a}{2}-b-c)^2+\frac{a^2}{12}+\frac{3}{a} > 0[/TEX]
[TEX]\Rightarrow (\frac{a}{2}-b-c)^2+\frac{a^3+36}{12a} > 0[/TEX]
Là BĐT đúng
2,
[TEX](a-1)^2+(b-1)^2+(a-b)^2 \geq 0[/TEX]
[TEX]\Rightarrow 2(a^2+b^2+1) \geq 2(ab+a+b)[/TEX]
[TEX]\Rightarrow a^2+b^2+1 \geq ab+a+b[/TEX]
3,
[TEX]\frac{1}{1+a} + \frac{1}{1+b} +\frac{1}{1+c}+\frac{1}{1+d} \geq 3[/TEX]
[TEX]\Rightarrow \frac{1}{1+a} \geq 1-\frac{1}{1+b}+1-\frac{1}{1+c}+1-\frac{1}{1+d}[/TEX]
[TEX]\Rightarrow \frac{1}{1+a} \geq \frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d} \geq 3\sqrt[]{\frac{bcd}{(b+1)(c+1)(d+1)}}[/TEX]
CM tương tự ta có:
[TEX]\Rightarrow \frac{1}{1+b} \geq \frac{b}{1+a}+\frac{c}{1+c}+\frac{d}{1+d} \geq 3\sqrt[]{\frac{acd}{(a+1)(c+1)(d+1)}}[/TEX]

[TEX]\Rightarrow \frac{1}{1+c} \geq \frac{a}{1+a}+\frac{b}{1+b}+\frac{d}{1+d} \geq 3\sqrt[]{\frac{abd}{(a+1)(b+1)(d+1)}}[/TEX]

[TEX]\Rightarrow \frac{1}{1+d} \geq \frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c} \geq 3\sqrt[]{\frac{abc}{(a+1)(b+1)(c+1)}}[/TEX]
Nhân lại với nhau ta có:
[TEX]\frac{1}{(a+1)(b+1)(c+1)(d+1)} \geq\frac{81abcd}{(a+1)b+1)(c+1)(d+1)}[/TEX]
[TEX]\Rightarrow abcd \leq \frac{1}{81}[/TEX]
 
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