Với `a;b;c>0` ta có:

*`(\sqrt{1/a}-1)^2\ge 0<=>1/a+1\ge 2. \sqrt{1/a}`

*`(\sqrt{a}-1)^2\ge 0<=>a+1\ge 2\sqrt{a}`

`=>(a+1).(1/a+1)\ge 2\sqrt{a}. 2 \sqrt{1/a}`

`<=>(a+1)(1/a+1)\ge 4`

`=>1/a+1\ge 4/{a+1}`

(Hoặc áp dụng luôn BĐT Cosi)

Tương tự:

 `1/b+1\ge 4/{b+1}`

`1/c+1\ge 4/{c+1}`

`=>1/a+1+1/b+1+1/c+1\ge 4/{a+1}+4/{b+1}+1/{c+1}`

`<=>1/a+1/b+1/c+3\ge 4(1/{a+1}+1/{b+1}+1/{c+1})`

Đẳng thức xảy ra khi $a=b=c=1$

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 $\begin{cases}a=1\\b=1\\c=1\end{cases} ⇔a=b=c=1.$