Cho x$\geq$ 1; y$\geq$ 1.  Chứng minh$\frac{1}{1+x^{2}}$ +$\frac{1}{1+y^{2}}$ $\geq$ $\frac{2}{1+xy}$ 

Giải thích các bước giải:

Xét 

$\begin{array}{l} \dfrac{1}{{1 + {x^2}}} + \dfrac{1}{{1 + {y^2}}} - \dfrac{2}{{1 + xy}}\\ = \dfrac{1}{{1 + {x^2}}} - \dfrac{1}{{1 + xy}} + \dfrac{1}{{1 + {y^2}}} - \dfrac{1}{{1 + xy}}\\ = \dfrac{{1 + xy - 1 - {x^2}}}{{\left( {1 + {x^2}} \right)\left( {1 + xy} \right)}} + \dfrac{{1 + xy - 1 - {y^2}}}{{\left( {1 + {y^2}} \right)\left( {1 + xy} \right)}}\\ = \dfrac{{xy - {x^2}}}{{\left( {1 + {x^2}} \right)\left( {1 + xy} \right)}} + \dfrac{{xy - {y^2}}}{{\left( {1 + {y^2}} \right)\left( {1 + xy} \right)}}\\ = \dfrac{{x(y - x)}}{{\left( {1 + {x^2}} \right)\left( {1 + xy} \right)}} + \dfrac{{y(x - y)}}{{\left( {1 + {y^2}} \right)\left( {1 + xy} \right)}}\\ = \dfrac{{x(y - x)\left( {1 + {y^2}} \right) + y(x - y)\left( {1 + {x^2}} \right)}}{{\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)\left( {1 + xy} \right)}}\\ = \dfrac{{ - (x - y)\left( {x + x{y^2}} \right) + (x - y)\left( {y + {x^2}y} \right)}}{{\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)\left( {1 + xy} \right)}}\\ = \dfrac{{(x - y)(y + {x^2}y - x - x{y^2})}}{{\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)\left( {1 + xy} \right)}}\\ = \dfrac{{(x - y)\left[ {\left( {{x^2}y - x{y^2}} \right) - (x - y)} \right]}}{{\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)\left( {1 + xy} \right)}}\\ = \dfrac{{(x - y)\left[ {xy\left( {x - y} \right) - (x - y)} \right]}}{{\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)\left( {1 + xy} \right)}}\\ = \dfrac{{(x - y)\left[ {(x - y)(xy - 1)} \right]}}{{\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)\left( {1 + xy} \right)}}\\ = \dfrac{{{{(x - y)}^2}(xy - 1)}}{{\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)\left( {1 + xy} \right)}} \end{array}$ 

Vì `x≥1; y≥1 => 1+xy>0`

mà `1+x^2 >0; 1+y^2 >0` với mọi `x;y`

`=> (1+x^2)(1+y^2)(1+xy)>0`

Vì `x≥1; y≥1 => xy≥1  => xy-1≥0`

mà `(x-y)^2 ≥0` với mọi `x;y`

`=> (x-y)^2(xy-1)≥0`

`=>`  $\dfrac{{{{(x - y)}^2}(xy - 1)}}{{\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)\left( {1 + xy} \right)}} \ge 0$

Vậy $\dfrac{1}{{1 + {x^2}}} + \dfrac{1}{{1 + {y^2}}} - \dfrac{2}{{1 + xy}} \ge 0$

`=>` $\dfrac{1}{{1 + {x^2}}} + \dfrac{1}{{1 + {y^2}}} \ge \dfrac{2}{{1 + xy}}$