![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(a^2+b^2+c^2=2\)
\(\Leftrightarrow\left(a+b+c\right)^2-2ab-2ac-2bc=2\)
Mà a+b+c=2
\(\Rightarrow4-2ab-2ac-2bc=2\)
\(\Rightarrow2-2ab-2ac-2bc=0\)
\(\Rightarrow-2\left(ab+ac+bc\right)=-2\)
\(\Rightarrow ab+ac+bc=1\left(1\right)\)
Ta lại có:\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+ac+bc}{abc}\)
Từ (1) suy ra đc:\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\left(đpcm\right)\)
theo bài ra ta có: a+b+c=2 => (a+b+c)^2 =4 => a^2 +b^2 +c^2 +2(ab+bc+ca)=4=> 2(ab+bc+ca)=2(vì a^2 +b^2 +c^2=2)
=> ab+bc+ca=1 =>\(\frac{ab}{abc}+\frac{bc}{abc}+\frac{ca}{abc}=\frac{1}{abc}\) (vì abc khác 0)
=> \(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=\frac{1}{abc}\)
Vậy với a+b+c=a^2+b^2+c^2=2 và abc khác 0 thì \(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=\frac{1}{abc}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) = \(\overline{\frac{\overline{bc}+\overline{ac}+\overline{ac}}{\overline{abc}}}\) = ab + bc + ca
=> a + b + c = ab + bc + ca
=> a + b + c - ab - bc - ca = 0
=> a + b + c - ab - bc - ac + abc - 1 = 0
=> (a - ab) + (b - 1) + (c - bc) + (abc - ac) = 0
=> - a(b - 1) + (b - 1) - c(b - 1) + ac(b - 1) = 0
=> (b - 1)(- a + 1 - c + ac) = 0
=> (b - 1)[( - a + 1) + (ac - c)] = 0
=> (b - 1)[ - (a - 1) + c(a - 1)] = 0
=> (a - 1)(b - 1)(c - 1) = 0
=> a - 1 = 0 hoặc b - 1 = 0 hoặc c - 1 = 0
=> a = 1 hoặc b = 1 hoặc c = 1
Vậy (a - 1)(b - 1)(c - 1) > 1
\(\left(a-1\right)\left(b-1\right)\left(c-1\right)>0\)
\(\Leftrightarrow\left(ab-a-b+1\right)\left(c-1\right)>0\)
\(\Leftrightarrow abc-ac-bc+c-ab+a+b-1>0\)
\(\Leftrightarrow-ab-bc-ab+a+b+c>0\)
\(\Leftrightarrow a+b+c>ab+ac+bc\)
\(\Leftrightarrow a+b+c>\frac{abc}{a}+\frac{abc}{b}+\frac{abc}{c}\)
\(\Leftrightarrow a+b+c>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) (thỏa mãn đề bài)
Vậy \(\left(a-1\right)\left(b-1\right)\left(c-1\right)>0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge2\sqrt{a}.2\sqrt{b}.2\sqrt{c}=8\sqrt{abc}=8\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
giả sử \(a+\frac{1}{a}\ge2\)
vì a > 0 => \(a^2+1\ge2a\)
<=> \(a^2+1-2a\ge0\)
<=> \(\left(a-1\right)^2\ge0\)( luôn đúng vs mọi a > 0)
=> \(a+\frac{1}{a}\ge2\). CMTT ta có \(b+\frac{1}{b}\ge2\)và \(c+\frac{1}{c}\ge2\)(1)
Ta có \(\left(a+1\right)\left(b+1\right)\left(c+1\right)=abc+ac+bc+ab+a+b+c+1\)
\(=1+1+\frac{1}{b}+\frac{1}{a}+\frac{1}{c}+a+b+c\)\(=2+\left(\frac{1}{a}+a\right)+\left(\frac{1}{b}+b\right)+\left(\frac{1}{c}+c\right)\)
Từ (1) =>\(2+\left(\frac{1}{a}+a\right)+\left(\frac{1}{b}+b\right)+\left(\frac{1}{c}+c\right)\ge8\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A = \(\frac{a}{ab+a+1}\)\(+\)\(\frac{b}{bc+b+1}\)\(+\)\(\frac{c}{ac+c+1}\)
= \(\frac{a}{ab+a+1}\)\(+\)\(\frac{ab}{a\left(bc+b+1\right)}\)\(+\)\(\frac{abc}{ab\left(ac+c+1\right)}\)
= \(\frac{a}{ab+a+1}\)\(+\)\(\frac{ab}{abc+ab+a}\)\(+\)\(\frac{abc}{abc.a+abc+ab}\)
Vì abc = 1 nên:
A = \(\frac{a}{ab+a+1}\)\(+\)\(\frac{ab}{ab+a+1}\)\(+\)\(\frac{1}{ab+a+1}\)
= \(\frac{a+ab+1}{ab+a+1}\)= 1
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì a,b,c > 0 và a+b+c=1
=> 0 < a,b,c < 1
=> 1-a, 1-b, 1-c > 0
Áp dụng bất đẳng thức cô-si cho các số dương ta có:
\(VP=4\left(1-a\right)\left(1-b\right)\left(1-c\right)\le4\cdot\dfrac{\left[\left(1-a\right)+\left(1-c\right)\right]^2}{4}\cdot\left(1-b\right)\)
\(=\left(2-a-c\right)^2\left(1-b\right)\)
\(=\left[2\left(a+b+c\right)-a-c\right]^2\left(1-b\right)\)
\(=\left(a+2b+c\right)^2\left(1-b\right)=\left(b+1\right)^2\left(1-b\right)=\left(b+1\right)\left(1-b^2\right)< b+1=a+2b+c=VT\)
Vậy VT > VP. Dấu "=" không xảy ra