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Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)
\(\Rightarrow2+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)
\(\Rightarrow2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=2\)
\(\Rightarrow\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=1\)
\(\Rightarrow\frac{a+b+c}{abc}=1\Rightarrow a+b+c=abc\left(đpcm\right)\)
\(\Rightarrow\frac{a+b+c}{abc}=\frac{1}{9}\Leftrightarrow\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=\frac{2}{9}\)
Lại có \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=1\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=1\)
Vậy 1/a^2+1/b^2+1/c^2=1-2/9=7/9 ( Sê đài )
\(\frac{1}{\frac{1}{a}+\frac{1}{b}}+\frac{1}{\frac{1}{b}+\frac{1}{c}}+\frac{1}{\frac{1}{c}+\frac{1}{a}}\)\(=\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\)
Áp dụng bđt AM-GM cho 3 số thực dương a,b,c ta được:
\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{\left(a+b\right)^2}{4\left(a+b\right)}+\frac{\left(b+c\right)^2}{4\left(b+c\right)}+\frac{\left(c+a\right)^2}{4\left(c+a\right)}\)
\(\Rightarrow\frac{1}{\frac{1}{a}+\frac{1}{b}}+\frac{1}{\frac{1}{b}+\frac{1}{c}}+\frac{1}{\frac{1}{c}+\frac{1}{a}}\le\frac{a+b+c}{2}\left(1\right)\)
Áp dụng bđt Cauchy-Schwarz dạng engel ta có:
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{1}{\frac{1}{a}+\frac{1}{b}}+\frac{1}{\frac{1}{b}+\frac{1}{c}}+\frac{1}{\frac{1}{c}+\frac{1}{a}}\le\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\left(đpcm\right)\)
\(\)
\(a+b+c=abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
\(\Leftrightarrow\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=2\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=9\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=9\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=9\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=7\)
Ta có
a + b + c = abc
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Ta lại có
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
Ta có:a+b+c=abc
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Ta lại có :\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
Anh/chị tham khảo ở đây nhé :
đặt x=a^2 + 2bc, y=b^2 + 2ac, z=c^2 + 2ab
=> x + y + z = (a + b + c)^2 <(=) 1
VT bpt : 1/x + 1/y + 1/z >(=) 3.căn3(1/xyz)...dùng cô-si cho 3 số nhé
mà x + y + z >(=) 3.căn3(xyz) <(=) 1
<=> 1/( 3.căn3 (xyz) >(=) 1 (ở đây là đổi chiều bđt)
<=> 1/ căn3 (xyz) >(=) 3
=> VT: 1/x + 1/y + 1/z >(=) 3.3 = 9
từ giả thiết, ta có \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
ta có \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\left(vi:\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\right)\) (ĐPCM)
^_^
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=4\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)
=>\(2+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)
=>\(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\)
=>\(\frac{c+a+b}{abc}=1\)
=> a+b+c=abc (đpcm)
Từ \(\left(1\right)\) suy ra : \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)
Do \(\left(2\right)\) nên \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1,\) suy ra \(\frac{a+b+c}{abc}=1\\.\)
Do đó \(a+b+c=abc\)