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đặt \(\left(a;b;c\right)=\left(\sqrt{\frac{yz}{x}};\sqrt{\frac{zx}{y}};\sqrt{\frac{xy}{z}}\right)\)\(\Rightarrow\)\(a^2+b^2+c^2=1\)
\(A=\Sigma\frac{1}{1-ab}=\Sigma\frac{2ab}{2\left(a^2+b^2+c^2\right)-2ab}+3\le\frac{1}{2}\Sigma\frac{\left(a+b\right)^2}{b^2+c^2+c^2+a^2}\)
\(\le\frac{1}{2}\Sigma\left(\frac{a^2}{c^2+a^2}+\frac{b^2}{b^2+c^2}\right)=\frac{9}{2}\)
dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=\frac{1}{3}\)
Theo giả thiết ta có : \(x+yz=yz-z-1=\left(z-1\right)\left(y+1\right)=\left(x+y\right)\left(y+1\right)\)
Tương tự : \(y+zx=\left(x+y\right)\left(x+1\right)\)
Và \(z+xy=\left(x+1\right)\left(y+1\right)\)
Nên \(P=\frac{x}{\left(x+y\right)\left(y+1\right)}+\frac{y}{\left(x+y\right)\left(x+1\right)}+\frac{z^2+2}{\left(x+1\right)\left(y+1\right)}\)
\(=\frac{x^2+y^2+x+y}{\left(x+y\right)\left(x+1\right)\left(y+1\right)}+\frac{z^2+2}{\left(x+1\right)\left(y+1\right)}\)
Ta có \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2};\left(x+1\right)\left(y+1\right)\le\frac{\left(x+y+2\right)^2}{4}\)
nên \(P\ge\frac{2\left(x+y\right)^2+4\left(x+y\right)}{\left(x+y+2\right)^2\left(x+y\right)}+\frac{4\left(z^2+2\right)}{\left(x+y+2\right)^2}=\frac{2\left(x+y\right)+4}{\left(x+y+2\right)^2}+\frac{4\left(z^2+2\right)}{\left(x+y+2\right)^2}\)
\(=\frac{2}{z+1}+\frac{4\left(z^2+2\right)}{\left(z+1\right)^2}=f\left(z\right);z>1\)
Lập bảng biến thiên ta được \(f\left(z\right)\ge\frac{13}{4}\) hay min \(P=\frac{13}{4}\) khi \(\begin{cases}z=3\\x=y=1\end{cases}\)
\(\sqrt{\frac{xy}{xy+z}}=\sqrt{\frac{xy}{xy+z\left(x+y+z\right)}}=\sqrt{\frac{xy}{\left(x+z\right)\left(y+z\right)}}\le\frac{1}{2}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)\)
Tương tự: \(\sqrt{\frac{yz}{yz+x}}\le\frac{1}{2}\left(\frac{y}{x+y}+\frac{z}{x+z}\right)\) ; \(\sqrt{\frac{zx}{zx+y}}\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{z}{y+z}\right)\)
Cộng vế với vế ta có đpcm
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
Bài này cũng dễ mà:
Áp dụng BĐT Cô-si, ta có:
\(y+z+1\ge3\sqrt[3]{yz}\)
\(\Rightarrow\)\(\dfrac{y+z+1}{3}\ge\sqrt[3]{yz}\)
\(\Rightarrow\)\(\dfrac{x}{\sqrt[3]{yz}}\ge\dfrac{3x}{y+z+1}\)
\(\Rightarrow\)\(\sum\dfrac{x}{\sqrt[3]{yz}}\ge\sum\dfrac{3x}{y+z+1}\)
Mà \(\sum\dfrac{3x}{y+z+1}=\sum\dfrac{3x^2}{xy+xz+x}\)
Áp dụng BĐT Cauchy -Schwaz:
\(\sum\dfrac{3x^2}{xy+xz+x}\ge\dfrac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\)
Mà:
\(xy+yz+xz\le x^2+y^2+z^2\)(BĐT phụ)
\(\Rightarrow\)\(2\left(xy+yz+xz\right)\le2\left(x^2+y^2+z^2\right)=6\)
Áp dụng BĐT Bunhicopski:
\(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=9\)
\(\Rightarrow x+y+z\le3\)
\(\Rightarrow2\left(xy+yz+xz\right)+x+y+z\le6+3=9\)
\(\Rightarrow\)\(\dfrac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{3\left(x+y+z\right)^2}{9}\ge\dfrac{\left(x+y+z\right)^2}{3}\ge xy+yz+xz\left(ĐPCM\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\)x=y=z=1
Lời giải:
Vì \(x,y,z\leq 1\Rightarrow (x-1)(y-1)(z-1)\leq 0\)
\(\Leftrightarrow (xy-x-y+1)(z-1)\leq 0\)
\(\Leftrightarrow x+y+z-xy-yz-xz+xyz-1\leq 0\)
\(\Leftrightarrow x+y+z-xy-yz-xz\leq 1-xyz\leq 1(*)\) (do \(xyz\geq 0\) )
Mặt khác:
\(y,z\in [0;1]\Rightarrow y^{2017}\leq y; z^{2018}\leq z\)
Do đó:
\(T=x+y^{2017}+z^{2018}-xy-yz-xz\leq x+y+z-xy-yz-xz(**)\)
Từ \((*);(**)\Rightarrow T\leq 1\) hay \(T_{\max}=1\)
Dấu bằng xảy ra khi \((x,y,z)=(1,1,0);(0,0,1)\) hoặc hoán vị các bộ số ấy