a)$A=\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)$

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

Ta chứng minh bđt:$\dfrac{a}{b}+\dfrac{b}{a}\ge2$(1)

$\Leftrightarrow\dfrac{a^2+b^2}{ab}\ge2$

$\Leftrightarrow a^2+b^2\ge2ab$

$\Leftrightarrow\left(a-b\right)^2\ge0$(luôn đúng)

Áp dụng$\Rightarrow A\ge1+2+1=4\left(\text{đ}pcm\right)$

b)$B=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}$

$B=\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}$

$B=\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)$

Áp dụng bđt (1)$\Rightarrow B\ge2+2+2=6\left(\text{đ}pcm\right)$