\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ab+ac}{abc}=2\)

\(\frac{bc+ab+ac}{a+b+c}=2\Leftrightarrow bc+ab+ac=2\left(a+b+c\right)\)

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{bc}+\frac{2}{ab}+\frac{2}{ac}\)( * )

Để \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)thì \(2\left(\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}\right)=2\Leftrightarrow\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}=1\)

\(\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}=\frac{a^2bc+bac^2+ab^2c}{\left(abc\right)^2}=\frac{abc\left(a+b+c\right)}{\left(abc\right)^2}=\frac{a+b+c}{abc}\)

mà a + b + c = abc \(\Rightarrow\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}=\frac{abc}{abc}=1\Leftrightarrow\frac{2}{bc}+\frac{2}{ab}+\frac{2}{ac}=2\)

thay \(\frac{2}{bc}+\frac{2}{ab}+\frac{2}{ac}=2\) vào ( * ) ta được \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2=2\left(đpcm\right)\)

\(\text{Ta có: }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=\frac{bc.ac+ab.ac+ab.bc}{ab.bc.ac}\)

\(=\frac{abc.\left(a+b+c\right)}{a^2b^2c^2}=\frac{a+b+c}{abc}=1\left(\text{vì }a+b+c=abc\right)\)

\(\text{Lại có: }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=2\text{ vì }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\text{ từ}\left(1\right)\)

Vậy ...

Từ giả thiết suy ra : 

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a+b+c}{abc}\right)=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)(vì a + b + c = abc)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2=2\left(đpcm\right)\)

bạn nhân ra hết cho mk

thanks bạn nhiều nha

Nhìn sơ qua hình như bài có thiếu đk gì đó...

\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)=4\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\)

Đặt \(\left(a^2;b^2;c^2\right)=\left(x^3;y^3;z^3\right)\) \(\Rightarrow xyz=1\)

Ta có BĐT quen thuộc: \(x^3+y^3\ge xy\left(x+y\right)\)

\(\Rightarrow P=\sum\frac{xyz}{x^3+y^3+xyz}\le\sum\frac{xyz}{xy\left(x+y\right)+xyz}=\sum\frac{z}{x+y+z}=1\)

\("="\Leftrightarrow a=b=c=1\)

Ai có cách nào khác với anh Nguyễn Việt Lâm không mọi người ?

Cho a = 1; b = 2; c= 1/2 suy ra suy ra VT = 101/126 < 1?

Agami Raito đề sai nha bạn, mình có đề khác cũng gần giống, bạn xem thử :

\(\frac{1}{a^3+b^3+1}+\frac{1}{b^3+c^3+1}+\frac{1}{a^3+c^3+1}\le1\)

Giả thiết như trên nhé

Giải theo đề anh Bonking nha!

Trước hết ta có BĐT sau \(x^3+y^3\ge xy\left(x+y\right)\Leftrightarrow\left(x+y\right)\left(x-y\right)^2\ge0\) với x, y > 0 áp dụng vào và ta có:

\(VT\le\Sigma_{cyc}\frac{1}{ab\left(a+b\right)+1}=\Sigma_{cyc}\frac{abc}{ab\left(a+b+c\right)}=\frac{a+b+c}{a+b+c}=1\) (đpcm)

Dấu "=" xảy ra khi a = b = c

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

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

\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^2+2\left(\frac{1}{a}+\frac{1}{b}\right)\frac{1}{c}+\left(\frac{1}{c}\right)^2=4\)

\(\Leftrightarrow\left(\frac{1}{a}\right)^2+2\frac{1}{a}.\frac{1}{b}+\left(\frac{1}{b}\right)^2+2\left(\frac{1}{ac}+\frac{1}{bc}\right)+\left(\frac{1}{c}\right)^2=4\)

\(\Leftrightarrow\left(\frac{1}{a}\right)^2+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2+2\frac{1}{ab}+2\left(\frac{1}{ac}+\frac{1}{bc}\right)=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a}{abc}+\frac{b}{abc}+\frac{c}{abc}\right)=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a+b+c}{abc}\right)=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a+b+c}{a+b+c}\right)=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)

ok thank bn