\(5\left(x+y+z\right)^2\ge14\left(x^2+y^2+z^2\right)\)
\(\Leftrightarrow9x^2+9y^2+9z^2-10y\left(x+z\right)-10zx\le0\)
\(\Leftrightarrow9\left(\frac{x}{z}\right)^2+9\left(\frac{y}{z}\right)^2+9-10.\frac{y}{z}\left(\frac{x}{z}+1\right)-10\frac{x}{z}\le0\)
Đặt \(\left(\frac{x}{z};\frac{y}{z}\right)=\left(a;b\right)>0\)
\(9b^2-10b\left(a+1\right)+9a^2-10a+9\le0\)
Để BPT đã cho có nghiệm
\(\Rightarrow\Delta'=25\left(a+1\right)^2-9\left(9a^2-10a+9\right)\ge0\)
\(\Leftrightarrow25a^2+50a+25-81a^2+90a-81\ge0\)
\(\Leftrightarrow-56a^2+140a-56\ge0\Rightarrow\frac{1}{2}\le a\le2\)
\(P=\frac{2a+1}{a+2}\Rightarrow\frac{4}{5}\le P\le\frac{5}{4}\)
\(\Rightarrow P_{min}=\frac{4}{5}\) khi \(a=\frac{1}{2}\) hay \(z=2x\); \(P_{max}=\frac{5}{4}\) khi \(x=2z\)
Áp dụng BĐT Bunhiakowski ta có:
\(\left(x+y+z\right)^2\le\left[\frac{5}{9}\left(x+z\right)^2+y^2\right]\left(\frac{9}{5}+1\right)\)
\(\Leftrightarrow5\left(x+y+z\right)^2\le14\left[\frac{5}{9}\left(x+z\right)^2+y^2\right]\)
\(\Rightarrow14\left(x^2+y^2+z^2\right)\le14\left[\frac{5}{9}\left(x+z\right)^2+y^2\right]\)
\(\Leftrightarrow2x^2-5xz+2z^2\le0\)
\(\Leftrightarrow\left(2x-z\right)\left(x-2z\right)\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x\ge z;x\le2z\\2x\le z;x\ge2z\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x\ge z\\2z\ge x\end{matrix}\right.\)
\(\Leftrightarrow\frac{1}{2}\le\frac{x}{z}\le2\).
Đặt \(\frac{x}{z}=o>0\).
Ta có: \(P=\frac{2t+1}{t+2}=2-\frac{3}{t+2}\).
Mặt khác \(\frac{1}{2}\le t\le2\) nên \(\frac{4}{5}\le P\le\frac{5}{4}\).
Vậy Min P = \(\frac{4}{5}\) khi x = 1; y = \(\frac{5}{3}\); z = 2.
Max P = \(\frac{5}{4}\) khi x = 2; y = \(\frac{5}{3}\); z = 1.
