Áp dụng bđt Cô-si có \(\Sigma\left(\frac{x^2}{y+1}+\frac{y+1}{4}\right)\ge\Sigma2\sqrt{\frac{x^2}{y+1}.\frac{y+1}{4}}=\Sigma x\)
\(\Rightarrow\Sigma\frac{x^2}{y+1}+\Sigma\frac{y+1}{4}\ge\Sigma x\)
\(\Rightarrow\Sigma\frac{x^2}{y+1}\ge\frac{3\Sigma x}{4}-\frac{3}{4}\)
Theo bđt Cô-si \(\Sigma x\ge3\sqrt[3]{\Pi x}=3\)
\(\Rightarrow\Sigma\frac{x^2}{y+1}\ge\frac{3\Sigma x}{4}-\frac{3}{4}\ge\frac{3.3}{4}-\frac{3}{4}=\frac{6}{4}=\frac{3}{2}\)
Dấu "='' <=> x = y = z = 1
Ta có \(P=\frac{x^2}{y+1}+\frac{y^2}{z+1}+\frac{z^2}{x+1}\) \(\Rightarrow P+\frac{x+y+z+3}{4}=P+\frac{X+1}{4}+\frac{Y+1}{4}+\frac{Z+1}{4}\)
= \(\left(\frac{x^2}{y+1}+\frac{y+1}{4}\right)+\left(\frac{y^2}{z+1}+\frac{z+1}{4}\right)+\left(\frac{z^2}{x+1}+\frac{x+1}{4}\right)\)
Do các số trong ngoặc đều dương nên áp dụng BĐT Cô - Si, ta có :
\(\frac{x^2}{y+1}+\frac{y+1}{4}\ge2\sqrt{\frac{x^2}{y+1}.\frac{y+1}{4}}=x\)
Tương tự suy ra \(\frac{y^2}{z+1}+\frac{z+1}{4}\ge y;\frac{z^2}{x+1}+\frac{x+1}{4}\ge z\)
Vậy P + \(\frac{x+y+z+3}{4}\ge x+y+z\Rightarrow P\ge\frac{3x+3y+3z-3}{4}\left(1\right)\)
Ta có x, y, z > 0 nên theo BĐT Cô - Si, ta có : \(x+y+z\ge3\sqrt[3]{xyz}=3\left(2\right)\)
Từ (1), (2); ta có P \(\ge\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
Ta có: \(xyz=1\Leftrightarrow\sqrt[3]{xyz}=1\Leftrightarrow1\le\frac{x+y+z}{3}\)( BĐT AM-GM )
\(\Leftrightarrow x+y+z\ge3\)
Dấu " = " xảy ra <=> x=y=z=1
Đặt \(P=\frac{x^2}{y+1}+\frac{y^2}{z+1}+\frac{z^2}{x+1}\)
Áp dụng BĐT Cauchy-schwarz ta có: (link c/m BĐT Cauchy-schwarz Xem câu hỏi )
\(P\ge\frac{\left(x+y+z\right)^2}{x+y+z+3}\ge\frac{\left(x+y+z\right)^2}{x+y+z+x+y+z}=\frac{x+y+z}{2}\)
Dấu " = " xảy ra <=> x=y=z=1
Áp dụng BĐT AM-GM ta có:
\(P\ge\frac{3.\sqrt[3]{xyz}}{2}=\frac{3}{2}\)
Dấu " = " xảy ra <=> x=y=z=1
Vậy...