Cho a, b, c có tổng bằng 1 (a, b, c > 0). Chứng minh rằng: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\).
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![](https://rs.olm.vn/images/avt/0.png?1311)
cho 3 số dương a,b,c có tổng bằng 1
chứng minh rằng : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 : \(\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge2\Leftrightarrow\dfrac{a^2+b^2}{ab}-2\ge0\Leftrightarrow\dfrac{a^2-2ab+b^2}{ab}\ge0\)
\(cmt\) \(tương\) \(tự\) \(với\) : \(\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\) \(và\) \(\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\) \(đều\) \(\ge2\) \(như\) \(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\ge2\)
\(\Rightarrow\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\ge9\) \(hay\) \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(a=b=c=\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
áp dụng BĐT:
1/a +1/b+1/c>= 9/a+b+c mà a+b+c=1
=>1/a+1/b+1/c≥9
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}=9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(\Leftrightarrow a=b=c\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)>=3\cdot\sqrt[3]{abc}\cdot3\cdot\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}=9\cdot\sqrt[3]{a\cdot b\cdot c\cdot\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}=9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT\ge\dfrac{1}{\left(a^2+1\right)-1}+\dfrac{1}{\left(b^2+1\right)-1}+\dfrac{1}{\left(c^2+1\right)-1}+4-\dfrac{4}{ab+1}+4-\dfrac{4}{bc+1}+4-\dfrac{4}{ca+1}\)
\(VT\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{4}{ab+1}-\dfrac{4}{bc+1}-\dfrac{4}{ca+1}+12\)
Mặt khác \(a;b;c\ge1\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\Rightarrow ab+1\ge a+b\) (và tương tự...)
\(\Rightarrow VT\ge\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}-\dfrac{4}{a+b}-\dfrac{4}{b+c}-\dfrac{4}{c+a}+12\)
\(VT\ge\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}-\dfrac{4}{a+b}-\dfrac{4}{b+c}-\dfrac{4}{c+a}+1+1+1+9\)
\(VT\ge\left(\dfrac{2}{a+b}-1\right)^2+\left(\dfrac{2}{b+c}-1\right)^2+\left(\dfrac{2}{c+a}-1\right)^2+9\ge9\)
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