1.
\(1-\dfrac{1}{x+1}=\dfrac{x}{x+1}=\dfrac{1}{\dfrac{y}{2}+1}+\dfrac{1}{\dfrac{z}{3}+1}\ge\dfrac{2}{\sqrt{\left(\dfrac{y}{2}+1\right)\left(\dfrac{z}{3}+1\right)}}=\dfrac{2\sqrt{6}}{\sqrt{\left(y+2\right)\left(z+3\right)}}\)
Tương tự:
\(\dfrac{y}{2+y}\ge\dfrac{2\sqrt{3}}{\sqrt{\left(x+1\right)\left(z+3\right)}}\) ; \(\dfrac{z}{3+z}\ge\dfrac{2\sqrt{2}}{\sqrt{\left(x+1\right)\left(y+2\right)}}\)
Nhân vế với vế và rút gọn:
\(xyz\ge2\sqrt{6}.2\sqrt{3}.2\sqrt{2}=48\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(2;4;6\right)\)
2.
\(1-\dfrac{x}{x+1}=\dfrac{2y}{y+1}+\dfrac{3z}{z+1}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{y}{y+1}+\dfrac{y}{y+1}+\dfrac{z}{z+1}+\dfrac{z}{z+1}+\dfrac{z}{z+1}\ge5\sqrt[5]{\dfrac{y^2z^3}{\left(y+1\right)^2\left(z+1\right)^3}}\)
\(\Leftrightarrow\dfrac{1}{\left(1+x\right)^5}\ge\dfrac{5^5y^2z^3}{\left(y+1\right)^2\left(z+1\right)^3}\) (1)
Tương tự ta có:
\(\dfrac{1}{1+y}=\dfrac{x}{1+x}+\dfrac{y}{1+y}+\dfrac{3z}{1+z}\ge5\sqrt[5]{\dfrac{xyz^3}{\left(1+x\right)\left(1+y\right)\left(1+z\right)^3}}\)
\(\Leftrightarrow\dfrac{1}{\left(1+y\right)^5}\ge\dfrac{5^5xyz^3}{\left(1+x\right)\left(1+y\right)\left(1+z\right)^3}\Leftrightarrow\dfrac{1}{\left(1+y\right)^8}\ge\dfrac{5^{10}x^2y^2z^6}{\left(1+x\right)^2\left(1+z\right)^6}\) (2)
\(\dfrac{1}{1+z}\ge5\sqrt[5]{\dfrac{xy^2z^2}{\left(1+x\right)\left(1+y\right)^2\left(1+z\right)^2}}\Leftrightarrow\dfrac{1}{\left(1+z\right)^5}\ge\dfrac{5^5xy^2z^2}{\left(1+x\right)\left(1+y\right)^2\left(1+z\right)^2}\)
\(\Leftrightarrow\dfrac{1}{\left(1+z\right)^9}\ge\dfrac{5^{15}x^3y^6z^6}{\left(1+x\right)^3\left(1+y\right)^6}\) (3)
Nhân vế (1);(2);(3):
\(\dfrac{1}{\left(1+x\right)^5\left(1+y\right)^8\left(1+z\right)^9}\ge\dfrac{5^{30}.x^5y^{10}z^{15}}{\left(1+x\right)^5\left(1+y\right)^8\left(1+z\right)^9}\)
\(\Leftrightarrow xy^2z^3\le\dfrac{1}{5^6}\)
\(P_{max}=\dfrac{1}{5^6}\) khi \(x=y=z=\dfrac{1}{5}\)
