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

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

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

aps dụng BĐTcauchy-schwarz dạng engel ta có

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3^2}{a+b+c}\)

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

⇔ \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)(đpcm)

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 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

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

=>

Lời giải +HD chi tiết

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

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

\(A=\left(\dfrac{a+b+c}{a}\right)+\left(\dfrac{a+b+c}{b}\right)+\left(\dfrac{a+b+c}{c}\right)\) {nhân pp}
\(A=\left(\dfrac{a}{a}+\dfrac{b}{a}+\dfrac{c}{a}\right)+\left(\dfrac{a}{b}+\dfrac{b}{b}+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{c}{c}\right)\){tách nhỏ ra}

\(A=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\) ghép lại theo định hướng

\(\left\{{}\begin{matrix}\dfrac{a}{b}=x\\\dfrac{b}{c}=y\\\dfrac{a}{c}=z\end{matrix}\right.\) \(\Rightarrow A=3+\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)+\left(z+\dfrac{1}{z}\right)\) {đổi biến viêt cho gọn }

\(A=3+2.3+\left(\sqrt{x}-2+\sqrt{\dfrac{1}{x}}\right)+\left(\sqrt{y}-2+\sqrt{\dfrac{1}{y}}\right)+\left(\sqrt{z}-2+\sqrt{\dfrac{1}{z}}\right)\)

{định hướng ghép bp}

\(A=9+\left(\sqrt{x}-\sqrt{\dfrac{1}{x}}\right)^2+\left(\sqrt{y}-\sqrt{\dfrac{1}{y}}\right)^2+\left(\sqrt{z}-\sqrt{\dfrac{1}{z}}\right)^2\)

\(\sum\left(x-\dfrac{1}{x}\right)^2\ge0\Rightarrow9+\sum\left(x-\dfrac{1}{x}\right)^2\ge9\Rightarrow A\ge9\)Kết thúc

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

Áp dụng BĐT cauchy-schwarz dạng engel ta có:

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