Cho a, b, c \(\in\)R và a, b, c \(\ne\)0. CMR: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{4}{a+b+c}\ne0\)
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\(\frac{1}{c}=\frac{1}{2}(\frac{1}{a}+\frac{1}{b})\)
\(\Rightarrow\frac{1}{c}:\frac{1}{2}=\frac{1}{a}+\frac{1}{b}\)
\(\Rightarrow\frac{2}{c}=\frac{a}{ab}+\frac{b}{ab}\)
\(\Rightarrow\frac{2}{c}=\frac{a+b}{ab}\)
\(\Rightarrow2ab=(a+b)\cdot c\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow ab-bc=ac-ab\)
\(\Rightarrow b(a-c)=a(c-b)\)
\(\frac{a}{c}=\frac{a-c}{c-b}(đpcm)\)
\(\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{1}{2}.\left(\frac{a+b}{ab}\right)\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}\)
\(\Rightarrow ac+cb=2ab\Rightarrow ac-ab=-cb+ba\Rightarrow a.\left(c-b\right)=b.\left(a-c\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)
bn ghi sai đề kìa :v
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\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Leftrightarrow\frac{2}{c}=\frac{1}{a}+\frac{1}{b}\)
\(\Leftrightarrow\frac{2}{c}=\frac{a+b}{ab}\)
\(\Leftrightarrow2ab=c\left(a+b\right)\left(2\right)\)
Mà \(\frac{a}{b}=\frac{a-c}{c-b}\)
\(\Leftrightarrow ac-ab=ab-bc\)
\(\Leftrightarrow2ab=c\left(a+b\right)\left(1\right)\)
Nhận thấy ( 1 )=( 2 ) => đpcm
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Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
\(\Rightarrow VT=\frac{bk}{bk-b}=\frac{bk}{b\left(k-1\right)}=\frac{k}{k-1}\left(1\right)\)
\(\Rightarrow VP=\frac{c}{c-d}=\frac{dk}{dk-d}=\frac{dk}{d\left(k-1\right)}=\frac{k}{k-1}\left(2\right)\)
Từ (1) và (2) =>Đpcm
Theo bất đẳng thức \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{4}{a+b+c}\ge\frac{9}{a+b+c}-\frac{4}{a+b+c}\)\(=\frac{5}{a+b+c}\ne0\)\(\Rightarrowđpcm\)
k cho minh nha
Bất đẳng thức \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\) chỉ đúng với x, y, z dương.