a) Áp dụng BĐT AM-GM, ta có:
\(VT\ge\dfrac{4xy}{2}+\dfrac{x}{4}+\dfrac{y}{4}=xy+\dfrac{x}{4}+xy+\dfrac{y}{4}\ge2\sqrt{xy.\dfrac{y}{4}}+2\sqrt{xy.\dfrac{x}{4}}=x\sqrt{y}+y\sqrt{x}\)
Vậy ta có đpcm.
c)
ta có : \(\dfrac{1}{1+x}\ge1-\dfrac{1}{1+y}+1-\dfrac{1}{1+z}=\dfrac{y}{y+1}+\dfrac{z}{z+1}\)
Áp dụng bất đẳng thức AM - GM :
\(\Rightarrow\dfrac{1}{1+x}\ge\dfrac{y}{y+1}+\dfrac{z}{z+1}\ge2\sqrt{\dfrac{yz}{\left(y+1\right)\left(z+1\right)}}\)
tương tự ta cũng có
\(\dfrac{1}{1+y}\ge2\sqrt{\dfrac{xz}{\left(x+1\right)\left(z+1\right)}}\) ; \(\dfrac{1}{1+z}\ge2\sqrt{\dfrac{xy}{\left(x+1\right)\left(y+1\right)}}\)
nhân ba vế bất đẳng thức lại với nhau ta được
\(\dfrac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge8\sqrt{\dfrac{x^2y^2z^2}{\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2}}\)
\(\Rightarrow\dfrac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge\dfrac{8xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(\Rightarrow xyz\le\dfrac{1}{8}\) \(\Rightarrow Max\left(xyz\right)=\dfrac{1}{8}\) khi \(x=y=z=\dfrac{1}{2}\)
