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Có thể là tìm a;b;c hoặc chứng minh cái đề cho ra là để hack não người đọc đó bác
\(...\Leftrightarrow\dfrac{a+b+c-3x}{a}+\dfrac{a+b+c-3x}{b}+\dfrac{a+b+c-3x}{c}=\dfrac{54x-3\left(a+b+c\right)}{a+b+c}\)
\(\Leftrightarrow\left(a+b+c-3x\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{54x-3\left(a+b+c\right)}{a+b+c}\)
\(\Leftrightarrow a+b+c-3x=\dfrac{54x-3\left(a+b+c\right)}{a+b+c}.\dfrac{abc}{ab+bc+ca}\)
\(\Leftrightarrow a+b+c-3x=\dfrac{54xabc}{\left(a+b+c\right)\left(ab+bc+ca\right)}-\dfrac{3abc}{ab+bc+ca}\)
\(\Leftrightarrow x\left(\dfrac{54abc}{\left(a+b+c\right)\left(ab+bc+ca\right)}+3\right)=a+b+c+\dfrac{3abc}{ab+bc+ca}\)
\(\Leftrightarrow x=\dfrac{a+b+c+\dfrac{3abc}{ab+bc+ca}}{\dfrac{54abc}{\left(a+b+c\right)\left(ab+bc+ca\right)+3}}\).
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
=> bc+ac+ab=0
ta có
\(bc+ac=-ab\)
<=> \(\left(bc+ac\right)^2=a^2b^2\)
<=> \(b^2c^2+a^2c^2+2abc^2=a^2b^2\)
<=> \(b^2c^2+a^2c^2-a^2b^2=-2abc^2\)
tương tự
\(a^2b^2+b^2c^2-c^2a^2=-2ab^2c\)
\(c^2a^2+a^2b^2-b^2c^2=-2a^2bc\)
thay vào E ta đc
\(E=\dfrac{-a^2b^2c^2}{2ab^2c}-\dfrac{a^2b^2c^2}{2abc^2}-\dfrac{a^2b^2c^2}{2a^2bc}\)
=\(-\dfrac{ac}{2}-\dfrac{ab}{2}-\dfrac{bc}{2}=\dfrac{-\left(ac+ab+bc\right)}{2}=0\) (vì ac+bc+ab=0 cmt)
Hình như sai đề :
Ta có : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=0\)
\(\Leftrightarrow\dfrac{ab+ac+bc}{abc}=0\)
\(\Leftrightarrow ab+ac+bc=0\) ( do \(a;b;c\ne0\) ) ( 1 )
Từ ( 1 ) \(\Rightarrow ab+bc=-ac\)
\(\Rightarrow\left(ab+bc\right)^2=\left[-\left(ac\right)\right]^2\)
\(\Rightarrow a^2b^2+b^2c^2+2ab^2c=a^2c^2\) ( * )
CMTT , ta được : \(\left\{{}\begin{matrix}b^2c^2+c^2a^2+2bc^2a=a^2b^2\\c^2a^2+a^2b^2+2a^2cb=b^2c^2\end{matrix}\right.\) ( *' )
Thay ( * ) và ( * ') vào E , ta được :
\(E=\dfrac{a^2b^2c^2}{a^2b^2+b^2c^2-\left(a^2b^2+b^2c^2+2b^2ac\right)}+\dfrac{a^2b^2c^2}{b^2c^2+c^2a^2-\left(b^2c^2+c^2a^2+2bc^2a\right)}\)
\(+\dfrac{a^2b^2c^2}{c^2a^2+a^2b^2-\left(c^2a^2+a^2b^2+2a^2cb\right)}\)
\(=\dfrac{a^2b^2c^2}{-2b^2ac}+\dfrac{a^2b^2c^2}{-2c^2ab}+\dfrac{a^2b^2c^2}{-2a^2cb}\)
\(=\dfrac{-ac}{2}+\dfrac{-ab}{2}+\dfrac{-bc}{2}\)
\(=\dfrac{-\left(ac+ab+bc\right)}{2}\)
\(=\dfrac{0}{2}=0\)
Vậy \(E=0\)
Ta có:
\(\dfrac{2a+b}{a+b}+\dfrac{2c+d}{c+d}+\dfrac{2b+c}{b+c}+\dfrac{2d+a}{d+a}=6\)
⇔ \(\left(\dfrac{2a+b}{a+b}-1\right)+\left(\dfrac{2c+d}{c+d}-1\right)+\left(\dfrac{2b+c}{b+c}-1\right)+\left(\dfrac{2d+a}{d+a}-1\right)=2\)
⇔ \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+d}+\dfrac{d}{d+a}=2\)
⇔ \(\left(1-\dfrac{a}{a+b}\right)-\dfrac{b}{b+c}+\left(1-\dfrac{c}{c+d}\right)-\dfrac{d}{d+a}=0\)
⇔ \(\dfrac{b}{a+b}-\dfrac{b}{b+c}+\dfrac{d}{c+d}-\dfrac{d}{d+a}=0\)
⇔ \(\dfrac{b\left(b+c\right)-b\left(a+b\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(d+a\right)-d\left(c+d\right)}{\left(c+d\right)\left(d+a\right)}=0\)
⇔ \(\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)
⇔ \(\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}-\dfrac{d\left(c-a\right)}{\left(c+d\right)\left(d+a\right)}=0\)
⇔ \(\left(c-a\right)\left(\dfrac{b}{\left(a+b\right)\left(b+c\right)}-\dfrac{d}{\left(c+d\right)\left(d+a\right)}\right)=0\)
⇒ \(\dfrac{b}{\left(a+b\right)\left(b+c\right)}-\dfrac{d}{\left(c+d\right)\left(d+a\right)}=0\) \(\left(a\ne c\right)\)
⇒ \(b\left(c+d\right)\left(d+a\right)-d\left(a+b\right)\left(b+c\right)=0\)
⇔ \(\left(bc+bd\right)\left(d+a\right)-\left(ad+bd\right)\left(b+c\right)=0\)
⇔ \(bcd+abc+bd^2+abd-abd-acd-b^2d-bcd=0\)
⇔ \(abc+bd^2-acd-b^2d=0\)
⇔ \(ac\left(b-d\right)-bd\left(b-d\right)=0\)
⇔ \(\left(b-d\right)\left(ac-bd\right)=0\)
⇒ \(ac-bd=0\) \(\left(b\ne d\right)\)
⇔ \(ac=bd\)
Khi đó:
\(A=abcd=\left(ac\right)^2\)
⇒ \(ĐPCM\)
\(\dfrac{a^3}{\left(a+2b\right)\left(b+2c\right)}+\dfrac{a+2b}{27}+\dfrac{b+2c}{27}\ge3\sqrt[3]{\dfrac{a^3\left(a+2b\right)\left(b+2c\right)}{27^2.\left(a+2b\right)\left(b+2c\right)}}=\dfrac{a}{3}\)
Tương tự:
\(\dfrac{b^3}{\left(b+2c\right)\left(c+2a\right)}+\dfrac{b+2c}{27}+\dfrac{c+2a}{27}\ge\dfrac{b}{3}\)
\(\dfrac{c^3}{\left(c+2a\right)\left(a+2b\right)}+\dfrac{c+2a}{27}+\dfrac{a+2b}{27}\ge\dfrac{c}{3}\)
Cộng vế:
\(VT+\dfrac{2\left(a+b+c\right)}{9}\ge\dfrac{a+b+c}{3}\)
\(\Rightarrow VT\ge\dfrac{a+b+c}{9}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
Áp dụng BĐT
\(\dfrac{9}{x+y+z}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\\ \Rightarrow\dfrac{9abc}{a+3a+2c}\\ =\dfrac{9}{\left(a+c\right)\left(b+c\right)+2b}\le\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{4}{2}\)
Tương tự với 2 BĐT còn lại rồi cộng vế theo vế
=> 9 vế trái
\(\le\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\\ +\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{a+b+c}{2}\\ =\dfrac{3\left(a+b+c\right)}{2}\\ \Rightarrow......._{\left(đpcm\right)}\)
