Áp dụng BĐT AM - GM, ta có \(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{1+y}{8}+\dfrac{1+z}{8}\ge3\sqrt[3]{\dfrac{x^3\left(1+y\right)\left(1+z\right)}{64\left(1+y\right)\left(1+z\right)}}=\dfrac{3}{4}x\)
Chứng minh tương tự, ta có \(A+\dfrac{1+x}{4}+\dfrac{1+y}{4}+\dfrac{1+z}{4}\ge\dfrac{3}{4}\left(x+y+z\right)\Rightarrow A\ge\dfrac{1}{2}\left(x+y+z\right)-\dfrac{3}{4}\)
Áp dụng BĐT AM - GM, ta có \(x+y+z\ge3\sqrt[3]{xyz}=3\)
Do đó \(A\ge\dfrac{3}{2}-\dfrac{3}{4}=\dfrac{3}{4}\left(đpcm\right)\)
Dấu = xảy ra <=> x = y = z = 1.
- Áp dụng bất đẳng thức Caushy, ta có:
\(x+y+z\ge\sqrt[3]{xyz}=\sqrt[3]{1}=3\).
\(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{1+y}{8}+\dfrac{1+z}{8}\ge\dfrac{3}{4}x\left(1\right)\)
\(\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{1+z}{8}+\dfrac{1+x}{8}\ge\dfrac{3}{4}y\left(2\right)\)
\(\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{8}+\dfrac{1+y}{8}\ge\dfrac{3}{4}z\left(3\right)\)
- Lấy \(\left(1\right)+\left(2\right)+\left(3\right)\), ta được:
\(A+\dfrac{1}{4}\left(x+y+z\right)+\dfrac{3}{4}\ge\dfrac{3}{4}\left(x+y+z\right)\)
\(\Leftrightarrow A\ge\dfrac{1}{2}\left(x+y+z\right)-\dfrac{3}{4}\)
- Mặt khác \(x+y+z\ge3\)
\(\Rightarrow A\ge\dfrac{1}{2}.3-\dfrac{3}{4}=\dfrac{3}{4}\left(đpcm\right)\)
- Dấu "=" xảy ra khi \(x=y=z=1\)
