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[congchuatuyet204]

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giúp vs nha ACE::-SS:-SS:-SS//@-)
1/
a,b,c>0 ,abc=1. cm
1/[a^3(b+c)] + 1/ [b^3 (c+a)] + 1/ [c^3 (a+b)] \geq 3/2

2/
a,b,c>0. a+b+c= 1.cm
2*(a/b+b/c+c/a) \geq(1+a)/(1-a) + (1+b)/1-b) + (c+1)/(1-c)
3/
a,b,c>0.cm
(a+b)^2 / (a^2+b^2+2c^2) + (b+c)^2 / (b^2+c^2+2a^2) +
(c+a)^2 / (c^2+a^2+2b^2) \leq 3
thank trc nha.

 

[bboy114crew]

giúp vs nha ACE::-SS:-SS:-SS//@-)
1/
a,b,c>0 ,abc=1. cm
1/[a^3(b+c)] + 1/[b^3 (c+a)] + 1/ [c^3 (a+b)] \geq 3/2

2/
a,b,c>0. a+b+c= 1.cm
2*(a/b+b/c+c/a) \geq(1+a)/(1-a) + (1+b)/1-b) + (c+1)/(1-c)
3/
a,b,c>0.cm
(a+b)^2 / (a^2+b^2+2c^2) + (b+c)^2 / (b^2+c^2+2a^2) +
(c+a)^2 / (c^2+a^2+2b^2) \leq 3
thank trc nha.

1/
ta có:
[TEX]\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} +\frac{1}{c^3(b+a)} [/TEX]
[TEX]=\frac{a^2b^2c^2}{a^3(b+c)} +\frac{a^2b^2c^2}{b^3(a+c)}+\frac{a^2b^2c^2}{c^3(b+a)} = \frac{(bc)^2}{ab+ac} + \frac{(ac)^2}{ab+bc}+ \frac{(ab)^2}{cb+ac} \geq \frac{(ab+bc+ac)^2}{2(ab+bc+ac)} = \frac{ab+bc+ac}{2}\geq \frac{3}{2}[/TEX]

 

[hoanggu95]

viết lại nè

2/
a,b,c>0 ;a+b+c=1.CM:
[TEX]2(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\geq \frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}[/TEX]
3/
a,b,c >0.CM:
[TEX]\sum \frac{{(a+b)}^{2}}{{a}^{2}+{b}^{2}+2{c}^{2}}\leq 3[/TEX]

 

[trongthanh95]

2/
a,b,c>0 ;a+b+c=1.CM:
[TEX]2(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\geq \frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}[/TEX]
3/
a,b,c >0.CM:
[TEX]\sum \frac{{(a+b)}^{2}}{{a}^{2}+{b}^{2}+2{c}^{2}}\leq 3[/TEX]

3> áp ụng trực tiếp cói svac
[TEX]\sum \frac{{(a+b)}^{2}}{{a}^{2}+{b}^{2}+2{c}^{2}} \leq \sum \frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}=3[/TEX]

 

[bboy114crew]

1/
cách 1:
ta có:
[TEX]\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} +\frac{1}{c^3(b+a)} [/TEX]
[TEX]=\frac{a^2b^2c^2}{a^3(b+c)} +\frac{a^2b^2c^2}{b^3(a+c)}+\frac{a^2b^2c^2}{c^3(b+a)} = \frac{(bc)^2}{ab+ac} + \frac{(ac)^2}{ab+bc}+ \frac{(ab)^2}{cb+ac} \geq \frac{(ab+bc+ac)^2}{2(ab+bc+ac)} = \frac{ab+bc+ac}{2}\geq \frac{3}{2}[/TEX]

cách 2:
Ap dụng BĐT cauchy- scharz,ta có:
[TEX][\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} +\frac{1}{c^3(b+a)}][a(b+c)+b(a+c)+c(a+b)] \geq (\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) = (ab+bc+ac)^2[/TEX]\Rightarrow[TEX]\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} +\frac{1}{c^3(b+a)} \geq \frac{1}{2}(ab+bc+ac) \geq \frac{3}{2} [/TEX]

 

[nokonyo]

2/
[TEX]a,b,c>0 ;a+b+c=1[/TEX].CM:
[TEX]2(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\geq \frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}[/TEX]

[TEX]\frac{2a}{b}+\frac{2b}{c}+\frac{2c}{a} \geq \frac{1-a+2a}{1-a}+\frac{1-b+2b}{1-b}+\frac{1-c+2c}{1-c}[/TEX]
[TEX]\Leftrightarrow \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{3}{2}[/TEX]
[TEX]\Leftrightarrow \frac{ac}{b(b+c)}+\frac{ab}{c(a+c)}+\frac{bc}{a(a+b)} \geq \frac{3}{2}[/TEX]
[TEX]\Leftrightarrow \frac{a^2c^2}{abc(b+c)}+\frac{a^2^2b}{abc(a+c)}+ \frac{b^2c^2}{abc(a+b)} \geq \frac{3}{2}[/TEX]
(đúng theo BDT CBS)