[SỐ HỌC 6] BÀI TẬP TOÁN NÂNG CAO - PHẦN 2 (CÓ ĐÁP ÁN)

Bài toán 9: Thực hiện phép tính:

1) $\left( {\dfrac{1}{{2011}} + \dfrac{2}{{2010}} + \dfrac{3}{{2009}} + 4} \right):\left( {\dfrac{1}{{2009}} + \dfrac{1}{{2010}} + \dfrac{1}{{2011}} + \dfrac{1}{{2012}}} \right)$
2) $ - \dfrac{1}{{90}} - \dfrac{1}{{72}} - \dfrac{1}{{56}} - \dfrac{1}{{42}} - \dfrac{1}{{30}} - \dfrac{1}{{20}} - \dfrac{1}{{12}} - \dfrac{1}{6} - \dfrac{1}{2}$
3) $\dfrac{3}{2} - \dfrac{5}{6} + \dfrac{7}{{12}} - \dfrac{9}{{20}} + \dfrac{{11}}{{30}} - \dfrac{{13}}{{42}} + \dfrac{{15}}{{56}} - \dfrac{{17}}{{72}}$
4) $\left( {1 + \dfrac{1}{{1\cdot3}}} \right)\left( {1 + \dfrac{1}{{2\cdot4}}} \right)\left( {1 + \dfrac{1}{{3\cdot5}}} \right) \ldots \left( {1 + \dfrac{1}{{17\cdot19}}} \right)\left( {1 + \dfrac{1}{{18\cdot20}}} \right)$

Giải.

1) $\left( {\dfrac{1}{{2011}} + \dfrac{2}{{2010}} + \dfrac{3}{{2009}} + 4} \right):\left( {\dfrac{1}{{2009}} + \dfrac{1}{{2010}} + \dfrac{1}{{2011}} + \dfrac{1}{{2012}}} \right)$
$ = \left( {\dfrac{1}{{2011}} + 1 + \dfrac{2}{{2010}} + 1 + \dfrac{3}{{2009}} + 1 + 1} \right):\left( {\dfrac{1}{{2009}} + \dfrac{1}{{2010}} + \dfrac{1}{{2011}} + \dfrac{1}{{2012}}} \right)$
= $\left( {\dfrac{1}{{2011}} + \dfrac{{2011}}{{2011}} + \dfrac{2}{{2010}} + \dfrac{{2010}}{{2010}} + \dfrac{3}{{2009}} + \dfrac{{2009}}{{2009}} + \dfrac{{2012}}{{2012}}} \right):$

$\left( {\dfrac{1}{{2009}} + \dfrac{1}{{2010}} + \dfrac{1}{{2011}} + \dfrac{1}{{2012}}} \right)$

= $\left( {\dfrac{{2012}}{{2011}} + \dfrac{{2012}}{{2010}} + \dfrac{{2012}}{{2009}} + \dfrac{{2012}}{{2012}}} \right):\left( {\dfrac{1}{{2009}} + \dfrac{1}{{2010}} + \dfrac{1}{{2011}} + \dfrac{1}{{2012}}} \right)$
= $2012\left( {\dfrac{1}{{2011}} + \dfrac{1}{{2010}} + \dfrac{1}{{2009}} + \dfrac{1}{{2012}}} \right):\left( {\dfrac{1}{{2009}} + \dfrac{1}{{2010}} + \dfrac{1}{{2011}} + \dfrac{1}{{2012}}} \right)$
= $2012$
2) $ - \dfrac{1}{{90}} - \dfrac{1}{{72}} - \dfrac{1}{{56}} - \dfrac{1}{{42}} - \dfrac{1}{{30}} - \dfrac{1}{{20}} - \dfrac{1}{{12}} - \dfrac{1}{6} - \dfrac{1}{2}$
= $ - \left( {\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{{12}} + \dfrac{1}{{20}} + \dfrac{1}{{30}} + \dfrac{1}{{42}} + \dfrac{1}{{56}} + \dfrac{1}{{72}} + \dfrac{1}{{90}}} \right)$
= $ - \left( {\dfrac{1}{{1\cdot2}} + \dfrac{1}{{2\cdot3}} + \dfrac{1}{{3\cdot4}} + \dfrac{1}{{4\cdot5}} + \dfrac{1}{{5\cdot6}} + \dfrac{1}{{6\cdot7}} + \dfrac{1}{{7\cdot8}} + \dfrac{1}{{8\cdot9}} + \dfrac{1}{{9\cdot10}}} \right)$
= $ - \left( {\dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} +  \cdots  + \dfrac{1}{8} - \dfrac{1}{9} + \dfrac{1}{9} - \dfrac{1}{{10}}} \right)$
= $ - \left( {\dfrac{1}{1} - \dfrac{1}{{10}}} \right)$
= $\dfrac{{ - 9}}{{10}}$
3) $\dfrac{3}{2} - \dfrac{5}{6} + \dfrac{7}{{12}} - \dfrac{9}{{20}} + \dfrac{{11}}{{30}} - \dfrac{{13}}{{42}} + \dfrac{{15}}{{56}} - \dfrac{{17}}{{72}}$
$ = \left( {1 + \dfrac{1}{2}} \right) - \left( {\dfrac{1}{2} + \dfrac{1}{3}} \right) + \left( {\dfrac{1}{3} + \dfrac{1}{4}} \right) - \left( {\dfrac{1}{4} + \dfrac{1}{5}} \right) + \left( {\dfrac{1}{5} + \dfrac{1}{6}} \right) - \left( {\dfrac{1}{6} + \dfrac{1}{7}} \right)+ $

$ + \left( {\dfrac{1}{7} + \dfrac{1}{8}} \right) - \left( {\dfrac{1}{8} + \dfrac{1}{9}} \right)$

$ = 1 + \dfrac{1}{2} - \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{4} - \dfrac{1}{4} - \dfrac{1}{5} + \dfrac{1}{5} + \dfrac{1}{6} - \dfrac{1}{6} - \dfrac{1}{7} + \dfrac{1}{7} + \dfrac{1}{8} - \dfrac{1}{8} - \dfrac{1}{9}$
$ = 1 - \dfrac{1}{9}$
$ = \dfrac{8}{9}$
4) $\left( {1 + \dfrac{1}{{1\cdot3}}} \right)\left( {1 + \dfrac{1}{{2\cdot4}}} \right)\left( {1 + \dfrac{1}{{3\cdot5}}} \right) \ldots \left( {1 + \dfrac{1}{{17\cdot19}}} \right)\left( {1 + \dfrac{1}{{18\cdot20}}} \right)$
= $\left( {\dfrac{{1\cdot3}}{{1\cdot3}} + \dfrac{1}{{1\cdot3}}} \right)\left( {\dfrac{{2\cdot4}}{{2\cdot4}} + \dfrac{1}{{2\cdot4}}} \right)\left( {\dfrac{{3\cdot5}}{{3\cdot5}} + \dfrac{1}{{3\cdot5}}} \right) \ldots \left( {\dfrac{{18\cdot20}}{{18\cdot20}} + \dfrac{1}{{18\cdot20}}} \right)$
= $\dfrac{4}{{1\cdot3}}\cdot\dfrac{9}{{2\cdot4}}\cdot\dfrac{{16}}{{3\cdot5}} \ldots \dfrac{{324}}{{17\cdot19}}\cdot\dfrac{{361}}{{18\cdot20}}$
= $\dfrac{{2\cdot2}}{{1\cdot3}}\cdot\dfrac{{3\cdot3}}{{2\cdot4}}\cdot\dfrac{{4\cdot4}}{{3\cdot5}} \ldots \dfrac{{18\cdot18}}{{17\cdot19}}\cdot\dfrac{{19\cdot19}}{{18\cdot20}}$
= $\left( {\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3} \ldots \dfrac{{18}}{{17}}\cdot\dfrac{{19}}{{18}}} \right)\cdot\left( {\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5} \ldots \dfrac{{18}}{{19}}\cdot\dfrac{{19}}{{20}}} \right)$
= $19\cdot\dfrac{2}{{20}}$
= $\dfrac{{19}}{{10}}$
Bài toán 10: Cho ba số $a$, $b$, $c$ thỏa mãn $abc = 105$ và $bc + b +1$ $ \ne $ $0$. Tính giá trị của biểu thức $S = \dfrac{{105}}{{abc + ab + a}} + \dfrac{b}{{bc + b + 1}} + \dfrac{a}{{ab + a + 105}}$.

Giải.

$S = \dfrac{{105}}{{abc + ab + a}} + \dfrac{b}{{bc + b + 1}} + \dfrac{a}{{ab + a + 105}}$
= $\dfrac{{abc}}{{abc + ab + a}} + \dfrac{b}{{bc + b + 1}} + \dfrac{a}{{ab + a + abc}}$
= $\dfrac{{abc}}{{a\left( {bc + b + 1} \right)}} + \dfrac{b}{{bc + b + 1}} + \dfrac{a}{{a\left( {b + 1 + bc} \right)}}$
= $\dfrac{{bc}}{{bc + b + 1}} + \dfrac{b}{{bc + b + 1}} + \dfrac{1}{{bc + b + 1}}$
= $\dfrac{{bc + b + 1}}{{bc + b + 1}}$
= $1$

Bài toán 11: Chứng tỏ rằng:

$\left( {1 + \dfrac{1}{3} + \dfrac{1}{5} +  \ldots  + \dfrac{1}{{101}}} \right) - \left( {\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} +  \ldots  + \dfrac{1}{{100}} + \dfrac{1}{{102}}} \right) = $

$ = \dfrac{1}{{52}} + \dfrac{1}{{53}} + \dfrac{1}{{54}} +  \ldots  + \dfrac{1}{{101}} + \dfrac{1}{{102}}$

Giải.

$\left( {1 + \dfrac{1}{3} + \dfrac{1}{5} +  \ldots  + \dfrac{1}{{101}}} \right) - \left( {\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} +  \ldots  + \dfrac{1}{{100}} + \dfrac{1}{{102}}} \right) $
$ = \left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} +  \ldots  + \dfrac{1}{{101}} + \dfrac{1}{{102}}} \right) - 2\left( {\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} +  \ldots  + \dfrac{1}{{100}} + \dfrac{1}{{102}}} \right)$
= $\left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} +  \ldots  + \dfrac{1}{{101}} + \dfrac{1}{{102}}} \right) - \left( {1 + \dfrac{1}{2} + \dfrac{1}{3} +  \ldots  + \dfrac{1}{{50}} + \dfrac{1}{{51}}} \right)$
= $\dfrac{1}{{52}} + \dfrac{1}{{53}} + \dfrac{1}{{54}} +  \ldots  + \dfrac{1}{{101}} + \dfrac{1}{{102}}$

Bài toán 12: Chứng tỏ rằng:
1) $\dfrac{1}{{1\cdot2}} + \dfrac{1}{{2\cdot3}} + \dfrac{1}{{3\cdot4}} +  \cdots  + \dfrac{1}{{n(n + 1)}} < 1$ ($n$ $ \in $ $\mathbb{N}$$^*$)
2) $\dfrac{1}{{{1^2}}} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} +  \ldots  + \dfrac{1}{{{n^2}}} < 2$ ($n$ $ \in $ $\mathbb{N}$, $n \ge 2$)
3) $\dfrac{{{{10}^8} + 2}}{{{{10}^8} - 1}} < \dfrac{{{{10}^8}}}{{{{10}^8} - 3}}$
4) $\dfrac{{{{10}^{2004}} + 1}}{{{{10}^{2005}} + 1}} > \dfrac{{{{10}^{2005}} + 1}}{{{{10}^{2006}} + 1}}$
5) $\dfrac{{2000}}{{2001}} + \dfrac{{2001}}{{2002}} > \dfrac{{2000 + 2001}}{{2001 + 2002}}$
6) $\dfrac{1}{{11}} + \dfrac{1}{{12}} + \dfrac{1}{{13}} + \dfrac{1}{{14}} + \dfrac{1}{{15}} + \dfrac{1}{{16}} + \dfrac{1}{{17}} + \dfrac{1}{{18}} + \dfrac{1}{{19}} + \dfrac{1}{{20}} > \dfrac{1}{2}$
7) $\dfrac{1}{5} + \dfrac{1}{{13}} + \dfrac{1}{{14}} + \dfrac{1}{{15}} + \dfrac{1}{{61}} + \dfrac{1}{{62}} + \dfrac{1}{{63}} < \dfrac{1}{2}$
8) $\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} +  \ldots  + \dfrac{1}{{63}} > 2$

Giải.

1) $\dfrac{1}{{1\cdot2}} + \dfrac{1}{{2\cdot3}} + \dfrac{1}{{3\cdot4}} +  \cdots  + \dfrac{1}{{n(n + 1)}} = \dfrac{n}{{n + 1}} < 1$ ($n$ $ \in $ $\mathbb{N}$$^*$)

2) Ta có: $\dfrac{1}{{{n^2}}} < \dfrac{1}{{(n - 1)n}} = \dfrac{1}{{n - 1}} - \dfrac{1}{n}$ ($n$ $ \in $ $\mathbb{N}$, $n \ge 2$)

$\dfrac{1}{{{1^2}}} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} +  \ldots  + \dfrac{1}{{{n^2}}} < 1 + \left( {\dfrac{1}{1} - \dfrac{1}{2}} \right) + \left( {\dfrac{1}{2} - \dfrac{1}{3}} \right) +  \ldots  + $

$ + \left( {\dfrac{1}{{n - 1}} - \dfrac{1}{n}} \right) = 2 - \dfrac{1}{n} < 2$

$ \Rightarrow $ $\dfrac{1}{{{1^2}}} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} +  \ldots  + \dfrac{1}{{{n^2}}} < 2$

3) $\dfrac{{{{10}^8} + 2}}{{{{10}^8} - 1}} < \dfrac{{{{10}^8}}}{{{{10}^8} - 3}}$

Ta có:

$\dfrac{{{{10}^8} + 2}}{{{{10}^8} - 1}} = \dfrac{{{{10}^8} - 1 + 3}}{{{{10}^8} - 1}} = 1 + \dfrac{3}{{{{10}^8} - 1}} = 1\dfrac{3}{{{{10}^8} - 1}}$

$\dfrac{{{{10}^8}}}{{{{10}^8} - 3}} = \dfrac{{{{10}^8} - 3 + 3}}{{{{10}^8} - 3}} = 1 + \dfrac{3}{{{{10}^8} - 3}} = 1\dfrac{3}{{{{10}^8} - 3}}$

Vì $\dfrac{3}{{{{10}^8} - 1}} < \dfrac{3}{{{{10}^8} - 3}}$

nên $\dfrac{{{{10}^8} + 2}}{{{{10}^8} - 1}} < \dfrac{{{{10}^8}}}{{{{10}^8} - 3}}$

4) $\dfrac{{{{10}^{2004}} + 1}}{{{{10}^{2005}} + 1}} > \dfrac{{{{10}^{2005}} + 1}}{{{{10}^{2006}} + 1}}$

Đặt $A = \dfrac{{{{10}^{2004}} + 1}}{{{{10}^{2005}} + 1}}$, $B = \dfrac{{{{10}^{2005}} + 1}}{{{{10}^{2006}} + 1}}$

$10A = \dfrac{{{{10}^{2004}} + 1}}{{{{10}^{2005}} + 1}}\cdot10 = \dfrac{{{{10}^{2005}} + 10}}{{{{10}^{2005}} + 1}} = 1 + \dfrac{9}{{{{10}^{2005}} + 1}}$

$10B = \dfrac{{{{10}^{2005}} + 1}}{{{{10}^{2006}} + 1}}\cdot10 = \dfrac{{{{10}^{2006}} + 10}}{{{{10}^{2006}} + 1}} = 1 + \dfrac{9}{{{{10}^{2006}} + 1}}$

Vì $\dfrac{9}{{{{10}^{2005}} + 1}} > \dfrac{9}{{{{10}^{2006}} + 1}}$

nên $10A > 10B$ (hay $A > B$)

Vậy $\dfrac{{{{10}^{2004}} + 1}}{{{{10}^{2005}} + 1}} > \dfrac{{{{10}^{2005}} + 1}}{{{{10}^{2006}} + 1}}$

5) $\dfrac{{2000}}{{2001}} + \dfrac{{2001}}{{2002}} > \dfrac{{2000 + 2001}}{{2001 + 2002}}$

Ta có:

$\dfrac{{2000}}{{2001}} > \dfrac{{2000}}{{2001 + 2002}}$    (1)

$\dfrac{{2001}}{{2002}} > \dfrac{{2001}}{{2001 + 2002}}$    (2)

(1), (2) $ \Rightarrow $ $\dfrac{{2000}}{{2001}} + \dfrac{{2001}}{{2002}} > \dfrac{{2000 + 2001}}{{2001 + 2002}}$

6) $\dfrac{1}{{11}} + \dfrac{1}{{12}} + \dfrac{1}{{13}} + \dfrac{1}{{14}} + \dfrac{1}{{15}} + \dfrac{1}{{16}} + \dfrac{1}{{17}} + \dfrac{1}{{18}} + \dfrac{1}{{19}} + \dfrac{1}{{20}} > \dfrac{1}{2}$

Ta có:

$\dfrac{1}{{11}} > \dfrac{1}{{20}}$

$\dfrac{1}{{12}} > \dfrac{1}{{20}}$

$.........$

$\dfrac{19}{{11}} > \dfrac{1}{{20}}$

Cộng theo vế ta được:

$\dfrac{1}{{11}} + \dfrac{1}{{12}} + \dfrac{1}{{13}} +  \ldots  + \dfrac{1}{{19}} > \underbrace {\dfrac{1}{{20}} + \dfrac{1}{{20}} + \dfrac{1}{{20}} +  \ldots  + \dfrac{1}{{20}}}_{{\rm{9 \text{ số} }}}$

$ \Rightarrow $ $\dfrac{1}{{11}} + \dfrac{1}{{12}} + \dfrac{1}{{13}} +  \ldots  + \dfrac{1}{{19}} + \dfrac{1}{{20}} > \underbrace {\dfrac{1}{{20}} + \dfrac{1}{{20}} + \dfrac{1}{{20}} +  \ldots  + \dfrac{1}{{20}} + \dfrac{1}{{20}}}_{{\rm{10 \text{ số}}}}$

$ \Rightarrow $ $\dfrac{1}{{11}} + \dfrac{1}{{12}} + \dfrac{1}{{13}} +  \ldots  + \dfrac{1}{{19}} + \dfrac{1}{{20}} > \dfrac{{10}}{{20}}$

$ \Rightarrow $ $\dfrac{1}{{11}} + \dfrac{1}{{12}} + \dfrac{1}{{13}} +  \ldots  + \dfrac{1}{{19}} + \dfrac{1}{{20}} > \dfrac{1}{2}$

7) $\dfrac{1}{5} + \dfrac{1}{{13}} + \dfrac{1}{{14}} + \dfrac{1}{{15}} + \dfrac{1}{{61}} + \dfrac{1}{{62}} + \dfrac{1}{{63}} < \dfrac{1}{2}$

Ta có:

$\dfrac{1}{5} + \dfrac{1}{{13}} + \dfrac{1}{{14}} + \dfrac{1}{{15}} + \dfrac{1}{{61}} + \dfrac{1}{{62}} + \dfrac{1}{{63}} < \dfrac{1}{5} + \dfrac{1}{{12}} + \dfrac{1}{{12}} + \dfrac{1}{{12}} + \dfrac{1}{{60}} + \dfrac{1}{{60}} + \dfrac{1}{{60}}$

$ \Rightarrow $ $S < \dfrac{1}{5} + \dfrac{1}{{12}}\cdot3 + \dfrac{1}{{60}}\cdot3 = \dfrac{1}{5} + \dfrac{1}{4} + \dfrac{1}{{20}} = \dfrac{4}{{20}} + \dfrac{5}{{20}} + \dfrac{1}{{20}} = \dfrac{{10}}{{20}}$

$ \Rightarrow $ $S < \dfrac{1}{2}$

8) $\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} +  \ldots  + \dfrac{1}{{63}} > 2$

Đặt $H = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} +  \ldots  + \dfrac{1}{{63}}$

$ \Rightarrow $ $H + 1 = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} +  \ldots  + \dfrac{1}{{63}}$

= $\left( {1 + \dfrac{1}{2}} \right) + \left( {\dfrac{1}{3} + \dfrac{1}{4}} \right) + \left( {\dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} + \dfrac{1}{8}} \right) + \left( {\dfrac{1}{9} + \dfrac{1}{{10}} +  \ldots  + \dfrac{1}{{16}}} \right) + $

$ + \left( {\dfrac{1}{{17}} + \dfrac{1}{{18}} +  \ldots  + \dfrac{1}{{32}}} \right) + \left( {\dfrac{1}{{33}} + \dfrac{1}{{34}} +  \ldots  + \dfrac{1}{{63}} + \dfrac{1}{{64}}} \right) - \dfrac{1}{{64}}$

$H + 1 > \dfrac{1}{2}\cdot2 + \dfrac{1}{4}\cdot2 + \dfrac{1}{8}\cdot4 + \dfrac{1}{{16}}\cdot8 + \dfrac{1}{{32}}\cdot16 + \dfrac{1}{{64}}\cdot32 - \dfrac{1}{{64}}$

$H + 1 > 1 + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{{64}}$

$H + 1 > 3 + \dfrac{{31}}{{64}}$

$H > 2 + \dfrac{{31}}{{64}} > 2$

Bài toán 13: Cho $a$, $b$, $c$, $d > 0$. Chứng tỏ rằng:

a) $\dfrac{{a + b}}{{1 + a + b}} \le \dfrac{a}{{1 + a}} + \dfrac{b}{{1 + b}}$

b) $1 < \dfrac{a}{{a + b + c}} + \dfrac{b}{{b + c + d}} + \dfrac{c}{{c + d + a}} + \dfrac{d}{{d + a + b}} < 2$

Giải.

a) Với $a, b > 0$ ta có:

$\dfrac{{a + b}}{{1 + a + b}} = \dfrac{a}{{1 + a + b}} + \dfrac{b}{{1 + a + b}} \le \dfrac{a}{{1 + a}} + \dfrac{b}{{1 + b}}$

b) Với $a$, $b$, $c$, $d$ ta có:

$\dfrac{a}{{a + b + c + d}} < \dfrac{a}{{a + b + c}} < \dfrac{a}{{a + c}}$    (1)

$\dfrac{b}{{b + c + d + a}} < \dfrac{b}{{b + c + d}} < \dfrac{b}{{b + d}}$    (2)

$\dfrac{c}{{c + d + a + b}} < \dfrac{c}{{c + d + a}} < \dfrac{c}{{c + a}}$    (3)

$\dfrac{d}{{d + a + b + c}} < \dfrac{d}{{d + a + b}} < \dfrac{d}{{d + b}}$    (4)

(1), (2), (3), (4) $ \Rightarrow $ đpcm

Bài toán 14: Chứng minh rằng tổng của một phân số dương với số nghịch đảo của nó thì không nhỏ hơn 2.

Giải.

Gọi phân số dương là $\dfrac{a}{b}$.

Không mất tính tổng quát, giả sử:

$a > 0$, $b > 0$, $a \ge b$, $a = b + m$ ($m$ $ \in $ $\mathbb{Z}$, $m \ge 0$)

Ta có:

$\dfrac{a}{b} + \dfrac{b}{a} = \dfrac{{b + m}}{b} + \dfrac{b}{{b + m}} = 1 + \dfrac{m}{b} + \dfrac{b}{{b + m}} \ge $

$ \ge 1 + \dfrac{m}{{b + m}} + \dfrac{b}{{b + m}} = 1 + \dfrac{{m + b}}{{b + m}} = 2$

Vậy $\dfrac{a}{b} + \dfrac{b}{a} \ge 2$, dấu “=” xảy ra khi $a = b$ (hay $m = 0$).