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Ta có: \(a+b+c=abc\)
=>\(\frac{a+b+c}{abc}=1\)
=>\(\frac{a}{abc}+\frac{b}{abc}+\frac{c}{abc}=1\)
=>\(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Lại có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
=>\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=2^2\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
=>ĐPCM
À thấy rồi, làm nè :
Ta có 1/a^2 + 1/b^2 + 1/c^2
= (1/a + 1/b + 1/c)^2 - 2 (1/ab + 1/ac + 1/bc)
= 4 - 2 (c/abc + b/ abc + a/ abc)
= 4 - 2 (a+b+c)/abc
= 4 - 2abc / abc
= 4 - 2
= 2 (đpcm)
gt: a/(b+c) + b/(c+a) + c/(a+b) = 1
A = a²/(b+c) + b²/(c+a) + c²/(a+b) = a[a/(b+c)] + b[b/(c+a)] + c[c/(a+b)]
= a[a/(b+c) + 1 - 1] + b[b/(c+a) + 1 - 1] + c[c/(a+b) + 1 - 1]
= a.(a+b+c)/(b+c) -a + b.(a+b+c)/(c+a) - b + c.(a+b+c)/(a+b) - c
= (a+b+c)[a/(b+c) + b/(c+a) + c/(a+b)] - (a+b+c)
= (a+b+c) - (a+b+c) = 0
Ta có: \(\frac{b^2-c^2}{\left(a+b\right)\left(a+c\right)}=\frac{b^2-a^2}{\left(a+b\right)\left(a+c\right)}+\frac{a^2-c^2}{\left(a+b\right)\left(a+c\right)}=\frac{b-a}{a+c}+\frac{a-c}{a+b}\left(1\right)\)
Tương tự ta có:
\(\frac{c^2-a^2}{\left(b+c\right)\left(b+a\right)}=\frac{c-b}{a+b}+\frac{b-a}{b+c}\left(2\right)\)
\(\frac{a^2-b^2}{\left(c+a\right)\left(c+b\right)}=\frac{a-c}{c+b}+\frac{c-b}{c+a}\left(3\right)\)
(1)(2)(3) => ĐPCM
Ta có :
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)
\(\Leftrightarrow2+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}=1\)
\(\Leftrightarrow\frac{a+b+c}{abc}=1\Leftrightarrow a+b+c=abc\left(đpcm\right)\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ab+ac}{abc}=2\)
\(\frac{bc+ab+ac}{a+b+c}=2\Leftrightarrow bc+ab+ac=2\left(a+b+c\right)\)
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{bc}+\frac{2}{ab}+\frac{2}{ac}\)( * )
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, 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)
^_^
a) Vì x;y;z > 0 nên áp dụng bất đẳng thức Bunhiakovsky : \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\) , ta được :
\(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}\ge\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+2xy+2yz+2xz}\)
\(\Leftrightarrow\)\(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)
Vậy \(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}\ge1\left(ĐPCM\right)\)
b) Ta chứng minh bất đẳng thức phụ :\(\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)
\(\Leftrightarrow\left(a+b+c\right)^2-3\left(ab+bc+ac\right)\ge0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac-3ab-3ac-3bc\ge0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-ab-ac\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) ( luôn đúng )
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+ab+ac\right)\)
Vì a,b,c > 0 nên áp dụng bất đẳng thức Bunhiakovsky : \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\) , ta được :
\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\)
mà \(\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\ge\frac{3\left(ab+bc+ac\right)}{2\left(ab+bc+ac\right)}=\frac{3}{2}\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{3}{2}\)
Vậy \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{3}{2}\left(ĐPCM\right)\)
Lời giải:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Rightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)
\(\Leftrightarrow \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=\frac{4-2}{2}=1\) (do \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\) )
\(\Leftrightarrow \frac{a+b+c}{abc}=1\)
\(\Leftrightarrow a+b+c=abc\)
Do đó ta có đpcm.
nhưng cô ơi trong đề chỉ nói 1/a+1/b+1/c=2 chứ có phải 1/a^2+1/b^2+1/c^2=2 đâu cô?