Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\text{Ta có: }\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{c}{abc}+\frac{a}{abc}+\frac{b}{abc}\right)\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{a+b+c}{abc}\right)\)
Mà \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\text{ và }a+b+c=abc\)nên:
\(2^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{abc}{abc}\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2=2\)
Bài 1 :
a) Ta có : \(\left(1-a\right)\left(1-b\right)\left(1-c\right)=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Áp dụng bđt Cauchy : \(a+b\ge2\sqrt{ab}\) , \(b+c\ge2\sqrt{bc}\) , \(c+a\ge2\sqrt{ca}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\) hay \(\left(1-a\right)\left(1-b\right)\left(1-c\right)\ge8abc\)
a) Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ta có :
\(\frac{1}{a+b-c}+\frac{1}{b+c-a}\ge\frac{4}{a+b-c+b+c-a}=\frac{4}{2b}=\frac{2}{b}\)
Tương tự :
\(\frac{1}{b+c-a}+\frac{1}{c+a-b}\ge\frac{4}{b+c-a+c+a-b}=\frac{4}{2c}=\frac{2}{c}\)
\(\frac{1}{a+b-c}+\frac{1}{c+a-b}\ge\frac{4}{a+b-c+c+a-b}=\frac{4}{2a}=\frac{2}{a}\)
Cộng theo vế :
\(\Rightarrow2\left(\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\left(đpcm\right)\)
Đẳng thức xảy ra <=> a = b = c
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=2^2\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^2+2\left(\frac{1}{a}+\frac{1}{b}\right)\frac{1}{c}+\left(\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow\left(\frac{1}{a}\right)^2+2\frac{1}{a}.\frac{1}{b}+\left(\frac{1}{b}\right)^2+2\left(\frac{1}{ac}+\frac{1}{bc}\right)+\left(\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow\left(\frac{1}{a}\right)^2+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2+2\frac{1}{ab}+2\left(\frac{1}{ac}+\frac{1}{bc}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a}{abc}+\frac{b}{abc}+\frac{c}{abc}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a+b+c}{abc}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a+b+c}{a+b+c}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
Có : 1/a + 1/b + 1/c = 2
<=> ( 1/a + 1/b + 1/c )^2 = 4
<=> 1/a^2 + 1/b^2 + 1/c^2 + 2.(1/ab + 1/bc + 1/ca) = 4
<=> 1/a^2 + 1/b^2 + 1/c^2 = 4 - 2.(1/ab + 1/bc + 1/ca)
= 4 - 2.(a+b+c)/abc
= 4 - 2 = 2
=> ĐPCM
Tk mk nha
Ta có : a + b + c = abc
\(\frac{\Rightarrow\left(a+b+c\right)}{abc}=\frac{abc}{abc}\)
\(\Rightarrow\frac{1}{ac}+\frac{1}{bc}+\frac{1}{ab}=1\)
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=2^2\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
\(\text{Ta có: }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=\frac{a+b+c}{abc}=\frac{abc}{abc}=1\left(\text{vì }a+b+c=abc\right)\)
\(\text{Lại có: }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2.1=2\left(\text{ vì }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\right)\)
Vậy ...