Chứng minh đẳng thức phụ: $a+b+c=0$ thì:
$|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}|=\sqrt[]{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}$
thật vậy:
$\sqrt[]{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}=\sqrt[]{(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^2-2.(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca})}$
$=\sqrt[]{(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^2-2.(\dfrac{a+b+c}{abc})}$
$=\sqrt[]{(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^2}$ (vì $abc=0$)
$=|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}|$
Áp dụng ta có:
$B=\sqrt[]{\dfrac{1}{1^2}+\dfrac{1}{1^2}+\dfrac{1}{(-2)^2}}+\sqrt[]{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{(-3)^2}}+...+\sqrt[]{\dfrac{1}{1^2}+\dfrac{1}{99^2}+\dfrac{1}{(-100)^2}}$
$=|1+\dfrac{1}{1}+\dfrac{1}{-2}|+|1+\dfrac{1}{2}+\dfrac{1}{-3}|+...+|1+\dfrac{1}{99}+\dfrac{1}{-100}|$
$=1+1-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{99}-\dfrac{1}{100}$
$=100-\dfrac{1}{100}$
