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Ta có : \(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=2\Leftrightarrow\frac{1}{x+1}=\left(1-\frac{1}{y+1}\right)+\left(1-\frac{1}{z+1}\right)\Leftrightarrow\frac{1}{x+1}=\frac{y}{y+1}+\frac{z}{z+1}\)
Tương tự ta cũng có : \(\frac{1}{y+1}=\frac{z}{z+1}+\frac{x}{x+1}\) ; \(\frac{1}{z+1}=\frac{y}{y+1}+\frac{x}{x+1}\)
Áp dụng bất đẳng thức Cosi: \(\frac{1}{x+1}=\frac{y}{y+1}+\frac{z}{z+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}\)
\(\frac{1}{y+1}\ge2\sqrt{\frac{xz}{\left(x+1\right)\left(z+1\right)}}\) ; \(\frac{1}{z+1}\ge2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\left(3\right)\)
Nhân (1) , (2) , (3) theo vế được :\(\frac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge8\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}.\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}.\sqrt{\frac{xz}{\left(x+1\right)\left(z+1\right)}}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge\frac{8xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(\Rightarrow8xyz\le1\Leftrightarrow xyz\le\frac{1}{8}\)(đpcm)
\(\frac{1}{x+1}=1-\frac{1}{y+1}+1-\frac{1}{z+1}=\frac{y}{y+1}+\frac{z}{z+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}\)
Tương tụ co:
\(\hept{\begin{cases}\frac{1}{y+1}\ge2\sqrt{\frac{zx}{\left(z+1\right)\left(x+1\right)}}\\\frac{1}{z+1}\ge2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\end{cases}}\)
\(\Rightarrow\frac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge\frac{8xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(\Leftrightarrow xyz\le\frac{1}{8}\)
Từ giả thiết:\(x+y+z=xyz\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)
Đặt \(\frac{1}{x}=a,\frac{1}{y}=b,\frac{1}{z}=c\)\(\Rightarrow ab+bc+ca=1\)
Ta có:\(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}+\frac{1}{\sqrt{1+z^2}}\)\(=\sqrt{\frac{1}{1+x^2}}+\sqrt{\frac{1}{1+y^2}}+\sqrt{\frac{1}{1+z^2}}\)
\(=\sqrt{\frac{\frac{1}{x}}{\frac{1}{x}+x}}+\sqrt{\frac{\frac{1}{y}}{\frac{1}{y}+y}}+\sqrt{\frac{\frac{1}{z}}{\frac{1}{z}+z}}\)\(=\sqrt{\frac{a}{a+\frac{1}{a}}}+\sqrt{\frac{b}{b+\frac{1}{b}}}+\sqrt{\frac{c}{c+\frac{1}{c}}}\)
\(=\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\)
Đến đây:\(\frac{a}{\sqrt{a^2+1}}=\frac{a}{\sqrt{a^2+ab+bc+ca}}=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(=\sqrt{\frac{a}{a+b}.\frac{a}{a+c}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)\)
Tương tự:\(\frac{b}{\sqrt{b^2+1}}\le\frac{1}{2}\left(\frac{b}{b+a}+\frac{b}{b+c}\right);\frac{c}{\sqrt{c^2+1}}\le\frac{1}{2}\left(\frac{c}{c+a}+\frac{c}{c+b}\right)\)
Cộng 3 bất đẳng thức lại ta có điều phải chứng minh :))
Từ (gt) \(\Rightarrow\frac{1}{1+x}=\left(1-\frac{1}{1+y}\right)+\left(1-\frac{1}{1+z}\right)=\frac{y}{1+y}+\frac{z}{1+z}\ge2\sqrt{\frac{yz}{\left(1+y\right)\left(1+z\right)}}\)
Tương tự \(\hept{\begin{cases}\frac{1}{1+y}\ge2\sqrt{\frac{xz}{\left(1+x\right)\left(1+z\right)}}\\\frac{1}{1+z}\ge2\sqrt{\frac{xy}{\left(1+x\right)\left(1+y\right)}}\end{cases}}\)
\(\Rightarrow\frac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge8\sqrt{\frac{\left(xyz\right)^2}{\left[\left(1+x\right)\left(1+y\right)\left(1+z\right)\right]^2}}=\frac{8xyz}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)
\(\Rightarrow xyz\le\frac{1}{8}\)
Đặt \(\frac{1}{1+x}=a\);\(\frac{1}{1+y}=b\);\(\frac{1}{1+y}=c\). Lúc đó a + b + c = 1
Ta có: \(a=\frac{1}{1+x}\Rightarrow x=\frac{1-a}{a}=\frac{\left(a+b+c\right)-a}{a}=\frac{b+c}{a}\)(Do a + b + c = 1)
Tương tự ta có: \(y=\frac{c+a}{b};z=\frac{a+b}{c}\)
\(\sqrt{x}+\sqrt{y}+\sqrt{z}\le\frac{3}{2}\sqrt{xyz}\Leftrightarrow\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}}+\frac{1}{\sqrt{xy}}\le\frac{3}{2}\)
Ta đi chứng minh \(\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(a+b\right)\left(b+c\right)}}\)\(\le\frac{3}{2}\)
\(VT\le\frac{1}{2}\left(\frac{a}{a+c}+\frac{b}{b+c}+\frac{b}{a+b}+\frac{c}{a+c}+\frac{a}{a+b}+\frac{c}{b+c}\right)\)
\(=\frac{1}{2}.3=\frac{3}{2}\)*đúng*
Vậy \(\sqrt{x}+\sqrt{y}+\sqrt{z}\le\frac{3}{2}\sqrt{xyz}\)
Đẳng thức xảy ra khi x = y = z = 2
\(\frac{1}{x+1}=1-\frac{1}{y+1}+1-\frac{1}{z+1}=\frac{y}{y+1}+\frac{z}{z+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}\)
Tương tự: \(\frac{1}{y+1}\ge2\sqrt{\frac{zx}{\left(z+1\right)\left(x+1\right)}}\) ; \(\frac{1}{z+1}\ge2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\)
Nhân vế với vế:
\(\frac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge\frac{8xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(\Rightarrow xyz\le\frac{1}{8}\)
Đề bài ko đúng rồi
\(\frac{1}{x+1}=\left(1-\frac{1}{y+1}\right)+\left(1-\frac{1}{z+1}\right)=\frac{y}{y+1}+\frac{z}{z+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}\) (1)
Tương tự :
\(\frac{1}{y+1}\ge2\sqrt{\frac{xz}{\left(x+1\right)\left(z+1\right)}}\) (2)
\(\frac{1}{z+1}\ge2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\) (3)
từ (1) (2) và (3) => \(\frac{1}{x+1}\cdot\frac{1}{y+1}\cdot\frac{1}{z+1}\ge8\sqrt{\frac{x^2y^2z^2}{\left[\left(x+1\right)\left(y+1\right)\left(z+1\right)\right]^2}}\)
=> \(\frac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge8\cdot\frac{xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
=> \(1\ge8xyz\)
=> \(xyz\le\frac{1}{8}\)
Dấu '=' xảy ra khi x = y = z = 1/2