toán 9

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kieutrang97

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thienvamai

[TEX]a^2+b^2\geq 2ab[/TEX]
[TEX]=> a^3+ab^2\geq2a^2b[/TEX]
tương tự [TEX]b^3+a^2b\geq2ab^2 [/TEX]
[TEX] => a^3+b^3\geq a^2b+b^2a=ab(a+b)[/TEX]


[TEX]a^3+b^3+1\geq ab(a+b)+1= ab(a+b+c)[/TEX]

[TEX]\frac{1}{a^3+b^3+1}\leq\frac{1}{ab(a+b+c)}=\frac{c}{a+b+c} [/TEX]
cm tương tự => đpcm
 
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