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Lời giải:
Ta có:
\(\left\{\begin{matrix} x=by+cz\\ y=ax+cz\\ z=ax+by\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x-y=by-ax\\ z=ax+by\end{matrix}\right.\)
\(\Rightarrow x-y+z=2by\Rightarrow b=\frac{x+z-y}{2y}\)
Hoàn toàn tương tự ta nhận được:
\(a=\frac{y+z-x}{2x};c=\frac{x+y-z}{2z}\)
Suy ra:
\(\left\{\begin{matrix} a+1=\frac{x+y+z}{2x}\\ b+1=\frac{x+y+z}{2y}\\ c+1=\frac{x+y+z}{2z}\end{matrix}\right.\)
\(\Rightarrow \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2\) (ĐPCM)
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:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(1+1+1\right)^2}{z+y+z}=9=\dfrac{18}{2}>\dfrac{18}{xyz+2}\)
Do \(x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}\)
=> \(\left\{{}\begin{matrix}x+\dfrac{1}{y}=y+\dfrac{1}{z}\Leftrightarrow x-y=\dfrac{1}{z}-\dfrac{1}{y}\Leftrightarrow x-y=\dfrac{y-z}{yz}\\y+\dfrac{1}{z}=z+\dfrac{1}{x}\Leftrightarrow y-z=\dfrac{1}{x}-\dfrac{1}{z}\Leftrightarrow y-z=\dfrac{z-x}{xz}\\z+\dfrac{1}{x}=x+\dfrac{1}{y}\Leftrightarrow z-x=\dfrac{1}{y}-\dfrac{1}{x}\Leftrightarrow z-x=\dfrac{x-y}{xy}\end{matrix}\right.\)
=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)=\dfrac{\left(y-z\right)\left(z-x\right)\left(x-y\right)}{x^2y^2z^2}\)
<=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)x^2y^2z^2=\left(y-z\right)\left(z-x\right)\left(x-y\right)\)
<=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x^2y^2z^2-1\right)=0\)
=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)=0\) hoặc \(x^2y^2z^2-1=0\)
=> x=y=z hoặc xyz=1 hoặc xyz=-1