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Ta có :
\(A=\frac{1+x^2}{1+y+z^2}+\frac{1+y^2}{1+z+x^2}+\frac{1+z^2}{1+x+y^2}\)
\(\Rightarrow A=\frac{1+z+x^2}{1+y+z^2}+\frac{1+x+y^2}{1+z+x^2}+\frac{1+y+z^2}{1+x+y^2}\)
\(-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)
\(\Rightarrow A\ge3\sqrt[3]{\frac{1+z+x^2}{1+y+z^2}.\frac{1+x+y^2}{1+z+x^2}.\frac{1+y+z^2}{1+x+y^2}}\)
\(-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)
\(\Rightarrow A\ge3-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)
\(\Rightarrow A\ge3-\left(\frac{z}{y+2z}+\frac{x}{z+2x}+\frac{y}{x+2y}\right)\)
\(\Rightarrow A\ge3-\left(\frac{1}{2}-\frac{y}{2\left(y+2z\right)}+\frac{1}{2}-\frac{z}{2\left(z+2x\right)}+\frac{1}{2}-\frac{x}{2\left(x+2y\right)}\right)\)
\(\Rightarrow A\ge3-\frac{3}{2}+\frac{1}{2}\left(\frac{y}{y+2z}+\frac{z}{z+2x}+\frac{x}{x+2y}\right)\)
\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{y^2}{y^2+2yz}+\frac{z^2}{z^2+2xz}+\frac{x^2}{x^2+2xy}\right)\)
\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{\left(x+y+z\right)^2}{y^2+2yz+z^2+2xz+x^2+2xy}\right)\)
\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}\right)\)
\(\Rightarrow A\ge2\)
Dấu " = " xảy ra khi \(x=y=z=1\)
Ta có :
\(\frac{1+x^2}{1+y+z^2}+\frac{1+y^2}{1+z+x^2}+\frac{1+z^2}{1+x+y^2}\)
\(\Rightarrow A=\frac{1+z+x^2}{1+y+z^2}+\frac{1+x+y^2}{1+z+x^2}+\frac{1+y+z^2}{1+x+y^2}\)
\(-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)
\(\Rightarrow A\ge3\sqrt[3]{\frac{1+z+x^2}{1+y+z^2}.\frac{1+x+y^2}{1+z+x^2}.\frac{1+y+z^2}{1+x+y^2}}\)
\(-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)
\(\Rightarrow A\ge3-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)
\(\Rightarrow A\ge3-\left(\frac{z}{y+2z}+\frac{x}{z+2x}+\frac{y}{x+2y}\right)\)
\(\Rightarrow A\ge3-\left(\frac{1}{2}-\frac{y}{2\left(y+2z\right)}+\frac{1}{2}-\frac{z}{2\left(z+2x\right)}+\frac{1}{2}-\frac{x}{2\left(x+2y\right)}\right)\)
\(\Rightarrow A\ge3-\frac{3}{2}+\frac{1}{2}\left(\frac{y}{y+2z}+\frac{z}{z+2x}+\frac{x}{x+2y}\right)\)
\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{y^2}{y^2+2yz}+\frac{z^2}{z^2+2xz}+\frac{x^2}{x^2+2xy}\right)\)
\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{\left(x+y+z\right)^2}{y^2+2yz+z^2+2xz+x^2+2xy}\right)\)
\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}\right)\)
\(\Rightarrow A\ge2\)
Dấu " = " xảy ra khi x=y=z=1
a) Đặt \(\hept{\begin{cases}x+y-z=a\\y+z-x=b\\z+x-y=c\end{cases}\Rightarrow}x=\frac{a+c}{2};y=\frac{b+a}{2};z=\frac{c+b}{2}\)
Suy ra bất đẳng thức cần chứng minh tương đương với: \(\frac{a+b}{2}.\frac{b+c}{2}.\frac{c+a}{2}\ge abc\Leftrightarrow\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{8}\ge abc\)\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)
Áp dụng bất đẳng thức AM-GM: \(\hept{\begin{cases}a+b\ge2\sqrt{ab}\ge0\\b+c\ge2\sqrt{bc}\ge0\\c+a\ge2\sqrt{ca}\ge0\end{cases}\Rightarrow}\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8\sqrt{\left(abc\right)^2}=8abc\)
Vật bất đẳng thức được chứng minh
Dấu "=" xảy ra khi \(a=b=c\Leftrightarrow x=y=z\)
Sử dụng bđt \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\left(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\right)^2\ge3\left(\frac{xy}{z}.\frac{yz}{x}+\frac{yz}{x}.\frac{zx}{y}+\frac{zx}{y}.\frac{xy}{z}\right)=3\left(x^2+y^2+z^2\right)=3\)
\(\Rightarrow\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\ge\sqrt{3}\)