\(\frac{a-1}{3}=\frac{b+2}{5}\)và ab=-2
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9 tháng 8 2020
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
29 tháng 4 2019
\(1)\) \(\frac{1}{a^3+5}+\frac{1}{b^3+5}\le\frac{1}{3a+3}+\frac{1}{3b+3}=\frac{1}{3}\left(\frac{1}{a+1}+\frac{1}{b+1}\right)=\frac{1}{3}\left[\frac{a+b+2}{\left(a+1\right)\left(b+1\right)}\right]\)
\(=\frac{1}{3}\left(\frac{ab+a+b+1}{ab+a+b+1}-\frac{ab-1}{ab+a+b+1}\right)=\frac{1}{3}\left(1-\frac{0}{ab+a+b+1}\right)=\frac{1}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}ab=1\\a^3=1\\b^3=1\end{cases}\Leftrightarrow a=b=1}\)
31 tháng 12 2019
2) bđt \(\Leftrightarrow\)\(\left(a-b\right)^2\left(\frac{1}{a}+\frac{1}{b}\right)+\left(a+b-2\right)\left[\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\frac{7}{2}\right]\ge0\) (1)
(1) đúng do \(a+b\ge2\sqrt{ab}=2\)\(\Leftrightarrow\)\(a+b-2\ge0\)
Dấu "=" xảy ra khi a=b=1
TM
28 tháng 9 2017
\(A=\frac{1}{\left(a+b\right)^3}.\frac{a^3+b^3}{\left(ab\right)^3}+\frac{3}{\left(a+b\right)^4}.\frac{a^2+b^2}{\left(ab\right)^2}+\frac{6}{\left(a+b\right)^5}.\frac{a+b}{ab}\)
\(=\frac{1}{\left(a+b\right)^3}.\frac{a^3+b^3}{1^3}+\frac{3}{\left(a+b\right)^4}.\frac{a^2+b^2}{1^2}+\frac{6}{\left(a+b\right)^5}.\frac{a+b}{1}\)
\(=\frac{a^2-ab+b^2}{\left(a+b\right)^2}+\frac{3\left(a^2+b^2\right)}{\left(a+b\right)^4}+\frac{6}{\left(a+b\right)^4}\)\(=\frac{\left(a^3+b^3\right)\left(a+b\right)+3a^2+3b^2+6}{\left(a+b\right)^4}\)
\(=\frac{a^4+a^3b+ab^3+b^4+3a^2+3b^2+6}{a^4+4a^3b+6a^2b^2+4ab^3+b^4}\)\(=\frac{a^4+a^2.1+1.b^2+b^4+3a^2+3b^2+6}{a^4+4a^2.1+6.1^2+4b^2.1+b^4}\)
\(=\frac{a^4+4a^2+4b^2+b^4+6}{a^4+4a^2+6+4b^2+b^4}=1\)