[Toán 10] BDT

U

usernames123

V

vodichhocmai

[TEX]\huge \blue x=\sqrt{\frac{b}{a}}\righ xy z=1[/TEX]

chúng ta có :

[TEX]\huge \blue \sqrt{\frac{2}{x^2+1}}\le \frac{3}{2}.\frac{(x+1\)}{x^2+x+1} =\frac{3}{2}\[1-\frac{x^2}{x^2+x+1} \][/TEX]

Do đó chúng ta cần chứng minh .

[TEX]\huge \blue \sum_{cyclic}^{xy z=1}\frac{x^2}{x^2+x+1} \ge 1[/TEX]

[TEX]\huge \blue \Leftrightarrow\sum_{cyclic}^{mnk=1}\frac{m^4}{m^4+m^2kn+k^2n^2}\ge \frac{(m^2+n^2+k^2\)^2}{m^4+n^4+k^4+mnk(m+n+k)+m^2n^2+n^2k^2+k^2n^2}[/TEX]

Nó là hiển nhiên đúng đo [TEX]\huge \blue m^2n^2+n^2k^2+k^2n^2\ge mnk(m+n+k)[/TEX]
 
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