[Toán 9] Chứng minh bất đẳng thức

C

congchuaanhsang

Vì x,y\geq1 nên ta có:
$(x-y)^2$(xy-1)\geq0
\Leftrightarrow(x-y)(x-y)(xy-1)\geq0
\Leftrightarrow(x-y)[xy(x-y)-(x-y)]\geq0
\Leftrightarrow(x-y)(-x-$xy^2$+y+$x^2y$)\geq0
\Leftrightarrow(x-y)[-x(1+$y^2$)+y(1+$x^2$)]\geq0
\Leftrightarrowx(y-x)(1+$y^2$)+y(x-y)(1+$x^2$)\geq0
\Leftrightarrow$\frac{x(y-x)}{(1+x^2)(1+xy)}$+$\frac{y(x-y)}{(1+y^2)(1+xy)}$\geq0
(Chia cả 2 vế cho số dương (1+$x^2$)(1+$y^2$)(1+xy))
\Leftrightarrow$\frac{1+xy-1-x^2}{(1+x^2)(1+xy)}$+$\frac{1+xy-1-y^2}{(1+y^2)(1+xy)}$\geq0
\Leftrightarrow$\frac{1}{1+x^2}$-$\frac{1}{1+xy}$+$\frac{1}{1+y^2}$-$\frac{1}{1+xy}$\geq0
\Leftrightarrow$\frac{1}{1+x^2}$+$\frac{1}{1+y^2}$-$\frac{2}{1+xy}$\geq0
\Leftrightarrow$\frac{1}{1+x^2}$+$\frac{1}{1+y^2}$\geq$\frac{2}{1+xy}$
 
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T

tranvanhung7997

Ta có: $\dfrac{1}{1 + a^2} + \dfrac{1}{1 + b^2} - \dfrac{2}{ab + 1}$
$= (\dfrac{1}{1 + a^2} - \dfrac{1}{ab + 1}) + (\dfrac{1}{1 + b^2} - \dfrac{1}{ab + 1})$
$ = \dfrac{ab - a^2}{(1 + ab)(1 + a^2)} + \dfrac{b^2 - ab}{(1 + ab)(1 + b^2)}$
$ = \dfrac{b - a}{1 + ab}.(\dfrac{a}{1 + a^2} - \dfrac{b}{1 + b^2}$
$ = \dfrac{(b - a)(ab^2 - a^2b + a - b)}{(1 + a^2)(1 + b^2)(1 + ab)}$
$ = \dfrac{(b - a)^2(ab - 1)}{(1 + a^2)(1 + b^2)(1 + ab)} \ge 0$ ( Vì $a \ge 1 ; b \ge 1$)
 
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