CHÚC BẠN HỌC TỐT!!!
Trả lời:
Ta có:
$(x-y)^2 \geq 0$
$⇒(x+y)^2-4xy \geq 0$
$⇒(x+y)^2 \geq 4xy$
$⇒\dfrac{(x+y)^2}{xy(x+y)} \geq \dfrac{4xy}{xy(x+y)}$
$⇒\dfrac{x+y}{xy} \geq \dfrac{4}{x+y}$
$⇒\dfrac{1}{x}+\dfrac{1}{y} \geq \dfrac{4}{x+y}$
Áp dụng BĐT trên:
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3=\bigg{(}\dfrac{1}{a}+\dfrac{1}{1}\bigg{)}+\bigg{(}\dfrac{1}{b}+\dfrac{1}{1}\bigg{)}+\bigg{(}\dfrac{1}{c}+\dfrac{1}{1}\bigg{)}$
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3 \geq \dfrac{4}{a+1}+\dfrac{4}{b+1}+\dfrac{4}{c+1}$
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3 \geq 4.\bigg{(}\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\bigg{)}$ (ĐPCM).
Dấu "=" xảy ra khi $\begin{cases}a=1\\b=1\\c=1\end{cases} ⇔a=b=c=1.$
