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\(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\\\dfrac{1}{z}=c\end{matrix}\right.\) \(\dfrac{\Rightarrow1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=a+b+c=0\)
cơ bản \(\left(a+b+c\right)=0\Rightarrow a^3+b^3+c^3=3abc\)
\(\Rightarrow x.y.z\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=\dfrac{1}{abc}.\left(a^3+b^3+c^3\right)=\dfrac{1}{abc}\left(3abc\right)=3=>dpcm\Leftrightarrow dccm\)
Ta có:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(1+1+1\right)^2}{z+y+z}=9=\dfrac{18}{2}>\dfrac{18}{xyz+2}\)
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Áp dụng C-S dạng Engel \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{\left(1+1+1\right)^2}{x+y+z}=\frac{9}{3\sqrt[3]{xyz}}=\frac{3}{\sqrt[3]{xyz}}\)
Vậy chứng minh \(\frac{3}{\sqrt[3]{xyz}}>\frac{18}{xyz+2}\Leftrightarrow xyz-6\sqrt[3]{xyz}+2>0\)
Đặt \(t=\sqrt[3]{xyz}\le\frac{x+y+z}{3}=\frac{1}{3}\Rightarrow0< t\le\frac{1}{3}\)
Hàm số \(f\left(t\right)=t^3-6t+2\) nghịch biến trên (\(0;\frac{1}{3}\)]
\(f\left(t\right)\ge f\left(\frac{1}{3}\right)=\frac{1}{27}>0\) (ĐPCM)
Đầu tiên ta cm:\(a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+\left(-a-b\right)^3=3abc\)
\(\Leftrightarrow a^3+b^3-a^3-3a^2b-3ab^2-b^3=3abc\)
\(\Leftrightarrow-3a^2b-3ab^2=3abc\)
\(\Leftrightarrow-3ab\left(a+b\right)=3abc\)
\(\Leftrightarrow-3ab\cdot\left(-c\right)=3abc\)(đúng)
Áp dụng:\(\Rightarrow xyz\cdot\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=xyz\cdot\dfrac{3}{xyz}=3\left(đpcm\right)\)
1.
Áp dụng BĐT Cauchy-Schwarz:
\(\dfrac{a}{2a+a+b+c}=\dfrac{a}{25}.\dfrac{\left(2+3\right)^2}{2a+a+b+c}\le\dfrac{a}{25}\left(\dfrac{2^2}{2a}+\dfrac{3^2}{a+b+c}\right)=\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{a}{a+b+c}\)
Tương tự:
\(\dfrac{b}{3b+a+c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{b}{a+b+c}\)
\(\dfrac{c}{a+b+3c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{c}{a+b+c}\)
Cộng vế:
\(VT\le\dfrac{6}{25}+\dfrac{9}{25}.\dfrac{a+b+c}{a+b+c}=\dfrac{3}{5}\)
Dấu "=" xảy ra khi \(a=b=c\)
2.
Đặt \(\dfrac{x}{x-1}=a;\dfrac{y}{y-1}=b;\dfrac{z}{z-1}=c\)
Ta có: \(\dfrac{x}{x-1}=a\Rightarrow x=ax-a\Rightarrow a=x\left(a-1\right)\Rightarrow x=\dfrac{a}{a-1}\)
Tương tự ta có: \(y=\dfrac{b}{b-1}\) ; \(z=\dfrac{c}{c-1}\)
Biến đổi giả thiết:
\(xyz=1\Rightarrow\dfrac{abc}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}=1\)
\(\Rightarrow abc=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)
\(\Rightarrow ab+bc+ca=a+b+c-1\)
BĐT cần chứng minh trở thành:
\(a^2+b^2+c^2\ge1\)
\(\Leftrightarrow\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\ge1\)
\(\Leftrightarrow\left(a+b+c\right)^2-2\left(a+b+c-1\right)\ge1\)
\(\Leftrightarrow\left(a+b+c-1\right)^2\ge0\) (luôn đúng)
Ta có: \(\left(x+y\right)+z^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)=x^2+y^2+z^2\)
\(\Rightarrow xy+yz+xz=0\Rightarrow\dfrac{xy+yz+xz}{xyz}=0\)
Hay \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=\dfrac{-1}{z}\Rightarrow\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3=\left(-\dfrac{1}{z}\right)^3\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{3}{xy}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{1}{y^3}=\dfrac{-1}{z^3}\)hay \(\dfrac{1}{x^3}-\dfrac{3}{xyz}+\dfrac{1}{y^3}=\dfrac{-1}{z^3}\)
\(\Rightarrow\dfrac{1}{x^2}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\)
