Mình bổ sung một cách làm khác nhé.

Áp dụng BĐT Cô-si cho 3 số dương \(a,b,c\), ta có \(a+b+c\ge3\sqrt[3]{abc}\) \(\Rightarrow1\ge3\sqrt[3]{abc}\)      (1)

Áp dụng BĐT Cô-si cho 3 số dương \(\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\) ta có \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\)           (2)

Nhân theo vế của các BĐT (1) và (2), ta được \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{abc}}=9\) (đpcm)

Đẳng thức xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{3}\)

\(Ta\) có : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

\(=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\)

\(=1+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{a}{b}+\dfrac{c}{b}+1+\dfrac{a}{c}+\dfrac{b}{c}+1\)

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

\(Ta\) có : \(\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge2\Leftrightarrow\dfrac{a^2+b^2}{ab}-2\ge0\Leftrightarrow\dfrac{a^2-2ab+b^2}{ab}\ge0\)

\(cmt\) \(tương\) \(tự\) \(với\) : \(\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\) \(và\) \(\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\) \(đều\) \(\ge2\) \(như\) \(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\ge2\)

\(\Rightarrow\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\ge9\) \(hay\) \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)

Áp dụng bất đẳng thức Cauchy-Schwarz:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{1}=9\)

Dấu "=" xảy ra khi: \(a=b=c=\dfrac{1}{3}\)

Áp dụng BĐT Svacxo ta được

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{\left(a+b+c\right)}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}=9\)

Vậy BĐT được chứng minh

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq (1+1+1)^2\)

\(\Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 9\) (đpcm)

Dấu bằng xảy ra khi \(a=b=c=\frac{1}{3}\)

áp dụng BĐT:
1/a +1/b+1/c>= 9/a+b+c mà a+b+c=1

=>

Áp dụng BĐT AM - GM, ta có:

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(=1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\)

\(=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)

\(\ge3+2+2+2=9\)

Dấu "=" xảy ra khi a = b = c

Áp dụng bất đẳng thức \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\) có:

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{9\left(a+b+c\right)}{\left(a+b+c\right)}=9\)

Dấu " = " khi a = b = c

BDT

\(x+\dfrac{1}{x}=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)^2+2\ge2\)

nhân PP vào là ra

\(\left(a+b+c\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3+2+2+2=9\)

Theo BĐT Cauchy:

\(a+b+c\ge3\sqrt[3]{abc}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\)

\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{abc}}=9\)

Đẳng thức xảy ra \(\Leftrightarrow a=b=c\)

Áp dụng BĐT C-S, ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)

\(\Rightarrow VT\ge9\)

Đẳng thức xảy ra khi a=b=c

Cách 2:

Ta có:

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

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

\(=1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\)

\(=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\)

Áp dụng BĐT AM-GM, ta có: \(\left\{{}\begin{matrix}\dfrac{a}{b}+\dfrac{b}{a}\ge2\\\dfrac{b}{c}+\dfrac{c}{b}\ge2\\\dfrac{c}{a}+\dfrac{a}{c}\ge2\end{matrix}\right.\)

=> \(A\ge9\)

P/s: Nhìn hơi dài nhưng trình bày ra thì không quá dài đâu! Ở đây mình làm hơi cẩn thận ::)))

Áp dụng Bất đẳng thức Côsi:

\(\left(a+b+c\right)\ge3\sqrt[3]{abc}\)

\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{\dfrac{1}{a}.\dfrac{1}{b}.\dfrac{1}{c}}\)

\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{a}.\dfrac{1}{b}.\dfrac{1}{c}}\)

\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)

Vậy \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\left(đpcm\right)\)

P/s: Ủa, đề này lớp 8 à? Sao cô mình lại cho bọn mình làm cái này nhỉ? WTF?????

Nguyễn thị kiều linh làm ko hiểu

a,b,c là các số dương nên \(\left(a+b+c\right)>=3\cdot\sqrt[3]{abc}\)

\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)>=3\cdot\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}\)

Do đó: \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)>=3\cdot\sqrt[3]{abc}\cdot3\cdot\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}=9\cdot\sqrt[3]{a\cdot b\cdot c\cdot\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}=9\)

Áp dụng AM-GM

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3.\dfrac{1}{\sqrt[3]{abc}}=9\)

\(\rightarrow1.\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)

vậy ta có điều phải chứng minh

Dấu "=" \(a=b=c=\dfrac{1}{3}\)

Áp dụng svac-xơ:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=9\)

Dấu = xảy ra <=> \(a=b=c=\dfrac{1}{3}\)

C2: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\)

\(=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\)

\(\ge3+2+2+2=9\) (theo cosi)

Dấu = xảy ra <=>\(a=b=c=\dfrac{1}{3}\)