Hi anh trai, nhớ em là ai chứ :))
Áp dụng BĐT AM - GM: \(x+y+z\ge3\sqrt[3]{xyz}=3\)
\(P=\Sigma\dfrac{1}{\left(3x+1\right)\left(y+z\right)+x}\) \(=\Sigma\dfrac{1}{3x\left(y+z\right)+x+y+z}\)
\(\Rightarrow P\le\Sigma\dfrac{1}{3x\left(y+z\right)+3}\)
\(\Leftrightarrow3P\le\Sigma\dfrac{1}{x\left(y+z\right)+1}\)
Chia cả hai vế cho \(xyz=1\)
\(\Leftrightarrow3P\le\Sigma\dfrac{1}{\dfrac{1}{y}+\dfrac{1}{z}+1}\)
Đặt \(a=\sqrt[3]{\dfrac{1}{x^3}},b=\sqrt[3]{\dfrac{1}{y^3}},c=\sqrt[3]{\dfrac{1}{z^3}}\)
\(\Rightarrow a.b.c=1\)
\(\Rightarrow3P\le\Sigma\dfrac{1}{a^3+b^3+1}\)
Mặt khác: \(\left(a-b\right)^2\ge0\)
\(\Leftrightarrow a^2-ab+b^2\ge ab\)
Nhân cả hai vế cho \(a+b\)
\(\Leftrightarrow a^3+b^3\ge ab\left(a+b\right)\)
\(\Leftrightarrow a^3+b^3+1\ge ab\left(a+b\right)+1=ab\left(a+b\right)+abc\)
\(\Leftrightarrow a^3+b^3+1\ge ab\left(a+b+c\right)\)
\(\Leftrightarrow3P\le\Sigma\dfrac{1}{ab\left(a+b+c\right)}=1\)
\(\Leftrightarrow P\le\dfrac{1}{3}\)
Dấu ''='' xảy ra \(\Leftrightarrow x=y=z=1\)
