Đáp án:
$a)$
\[A=\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)=\dfrac{a+1}{a}.\dfrac{b+1}{b}\] mà $a+b=1\to A=\dfrac{a+a+b}{a}.\dfrac{b+a+b}{b}=\dfrac{2a+b}{a}.\dfrac{a+2b}{b}$ \[\to A=\dfrac{2a^2+4ab+ab+2b^2}{ab}=\dfrac{2\left(a^2+2ab+b^2\right)}{ab}+\dfrac{ab}{ab}\] \[\to A=\dfrac{2}{ab}+1\]
Vì $\left(a-b\right)^2\ge0$
$\Rightarrow a^2+b^2-2ab\ge0$
$\Rightarrow2ab\le a^2+b^2$
$\Rightarrow4ab\le\left(a+b\right)^2=1$
$\Rightarrow ab\le\dfrac{1}{4}$
$\Rightarrow\dfrac{2}{ab}\ge8$
$\Rightarrow\dfrac{2}{ab}+1\ge9$ $\to A\ge 9$
Dấu $"="$ xảy ra $⇔a=b=\dfrac{1}{2}$
$b)$
Áp dụng bất đẳng thức cosi cho 2 số thực không âm:
\[\left\{\begin{matrix} \dfrac{a^3}{b}+ab\ge 2\sqrt{\dfrac{a^3}{b}.ab}= 2a^2& \\ \dfrac{b^3}{c}+bc\ge 2\sqrt{\dfrac{b^3}{c}.bc}=2b^2& \\ \dfrac{c^3}{a}+ac\ge 2\sqrt{\dfrac{c^3}{a}.ac}=2c^2 & \end{matrix}\right.\] \[\to \dfrac{a^3}{b}+ab+\dfrac{b^3}{c}+bc+\dfrac{c^3}{a}+ac\ge 2(a^2+b^2+c^2)\ge 2(ab+bc+ca)\] \[\to \dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ca\]
