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\(P=\frac{1}{x}+\frac{1}{y}+xy^2+x^2y=\left(\frac{1}{16x}+xy^2\right)+\left(\frac{1}{16y}+x^2y\right)+\frac{15}{16}\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(\ge\frac{y}{2}+\frac{x}{2}+\frac{15}{16}.\frac{4}{x+y}\)
\(=\left(\frac{x+y}{2}+\frac{1}{2\left(x+y\right)}\right)+\frac{13}{4\left(x+y\right)}\)
\(\ge1+\frac{13}{4}=\frac{17}{4}\)
Dấu "=" xảy ra <=> x = y = 1/2
\(P=\left(2x+\frac{1}{x}\right)^2+\left(2y+\frac{1}{y}\right)^2\)
Ta có: \(2x+\frac{1}{x}\ge2\sqrt{2x+\frac{1}{x}}=2\sqrt{2}\)
\(\Rightarrow\left(2x+\frac{1}{x}\right)^2\ge8\)
\(\Rightarrow\left(2y+\frac{1}{y}\right)^2\ge8\)
Dấu \("="\) xảy ra \(\Leftrightarrow x=y=\pm\frac{1}{2}\)
Vậy \(P_{min}=16\Leftrightarrow x=y=\pm\frac{1}{2}\)
A= (x2 +2x +1) +(y2+2y+1) +(z2+2z+1) -3
=(x+1)2 +(y+1)2 +(z+1)2 -3 >/ -3
A min = -3 khi x =y=z = -1
\(Q=3xy\left(x+3y\right)-2xy\left(x+4y\right)-x^2\left(y-1\right)+y^2\left(1-x\right)+36\)\(\Leftrightarrow Q=3x^2y+9xy^2-2x^2y-8xy^2-x^2y+x^2+y^2-xy^2+36\)\(\Leftrightarrow Q=\left(3x^2y-2x^2y-x^2y\right)+\left(9xy^2-8xy^2-xy^2\right)+x^2+y^2+36\)\(\Leftrightarrow Q=x^2+y^2+36\ge36\forall x;y\)
Dấu " = " xảy ra
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=0\\y^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy Min Q là : \(36\Leftrightarrow x=y=0\)