Chung minh a+b+c>=3(\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)) voi a,b,c>0 va a+b+c=abc
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Chung minh a+b+c>= 3(\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)) voi a,b,c>0 va a+b+c=abc
cho a,b, c > hoac = 0 va a+b+c=1.chung minh
\(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}>3.5\)
2 cho a,b,c >0 . chung minh
\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}>hoac=3\)
2. Áp dụng bất đẳng thức Cô - si cho 3 số dương \(\frac{a}{b},\frac{b}{c},\frac{c}{a}\)ta có
\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\sqrt[3]{\frac{a}{b}.\frac{b}{c}.\frac{c}{a}}\)\(=3\)
Dấu "=" xảy ra <=> a = b = c
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cho 3 so a,b,c khac 0 va (a+b+c)^2=a^2+b^2+c^2 . chung minh \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3abc\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2\)
\(\Rightarrow2\left(ab+bc+ac\right)=0\)
\(\Rightarrow ab+bc+ac=0\)
\(\Rightarrow\frac{\left(a+b+c\right)}{abc}=0\)
\(\Rightarrow\frac{ab}{abc}+\frac{bc}{abc}+\frac{ac}{abc}=0\)
\(\Rightarrow\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=0\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{-1}{c}\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(\frac{-1}{c}\right)^3\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}.\left(-\frac{1}{c}\right)=0\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{3}{ab}=0\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\left(đpcm\right)\)
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\(\left(a+b+c\right)^2=a^2+b^2+c^2\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2\Rightarrow ab+bc+ac=0\)
\(\Rightarrow\frac{ab+bc+ac}{abc}=0\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\left(\frac{1}{a}\right)^3+\left(\frac{1}{b}\right)^3+\left(\frac{1}{c}\right)^3=3.\frac{1}{a}.\frac{1}{b}.\frac{1}{c}\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
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cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\) va\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2.\)
Chung minh rang a+b+c=abc
Ta có:\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)
\(\Rightarrow2+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\Rightarrow\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=2\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
\(\Rightarrow\frac{a}{abc}+\frac{b}{abc}+\frac{c}{abc}=1\Rightarrow\frac{a+b+c}{abc}=1\Rightarrow a+b+c=abc\)
\(\Rightarrowđpcm\)
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Ta có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{2}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(\Rightarrow2^2=2+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(\Leftrightarrow2=.2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\)
\(\Leftrightarrow\frac{a}{abc}+\frac{a}{abc}+\frac{b}{abc}=\frac{abc}{abc}\)
\(\Leftrightarrow a+b+c=abc\)
\(\RightarrowĐPCM\)
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\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)
=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\)
=> \(\frac{a+b+c}{abc}=1\)
=> a+b+c=abc
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CMR \(\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)>=\left(\frac{10}{3}\right)^2\) voi a,b,c >0 va a+b+c=1
cho 1/a+1/b+1/c=2 va :a+b+c=abc .chung minh rang: \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
cho 1/a+1/b+1/c=2 va :a+b+c=abc
.chung minh rang:
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Giai va bien luan cac phuong trinh sau:
1. \(\frac{a+b-x}{c}+\frac{a+c-x}{b}+\frac{b+c-x}{a}+\frac{4x}{a+b+c}=1\)
(an x) voi dk; a,b,b khac 0 va a+b+c khac 0
2.\(\frac{x-a}{bc}+\frac{x-b}{ac}+\frac{x-c}{ab}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
(an x) voi dk: a,b,c khac 0
3, \(\frac{mx+3}{6}+\frac{m^2-1}{2}=\frac{x+5}{10}+\frac{2}{5}\left(x+m^2+1\right)\)
(an x)
Cho a;b;c>0
va ab+ac+bc=3
chung minh:\(\frac{1}{^a^{^2}+b^2+1}+\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}\le1\)
Cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\) (abc khác 0,a+b+c khác 0).Chứng minh \(\frac{1}{a^3}+\frac{1}{b^3^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^#}}}}}}}}}}}}}}}}+\frac{1}{c^3}=\frac{1}{a^3+b^3+c^3}\)
Chứng minh \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^3+b^3+c^3}\)