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\(2=3\sqrt{xy}+2\sqrt{xz}\le\dfrac{3}{2}\left(x+y\right)+x+z\)
\(\Rightarrow5x+3y+2z\ge4\)
\(A=5\left(\dfrac{xy}{z}+\dfrac{xz}{y}\right)+3\left(\dfrac{xy}{z}+\dfrac{yz}{x}\right)+2\left(\dfrac{xz}{y}+\dfrac{yz}{x}\right)\)
\(A\ge5.2x+3.2y+2.2z=2\left(5x+3y+2z\right)\ge8\)
\(A_{min}=8\) khi \(x=y=z=\dfrac{2}{5}\)
\(\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)
\(\Rightarrow2\ge3x^2+2y^2+2z^2+y^2+z^2\)
\(\Leftrightarrow2\ge3\left(x^2+y^2+z^2\right)\)
Có: \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)\le2\)
\(\Rightarrow\)\(A^2\le2\) \(\Leftrightarrow A\in\left[-\sqrt{2};\sqrt{2}\right]\)
minA=-1\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+y+z=-\sqrt{2}\\x=y=z\end{matrix}\right.\) \(\Rightarrow x=y=z=-\dfrac{\sqrt{2}}{3}\)
maxA=1\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=\sqrt{2}\\x=y=z\end{matrix}\right.\) \(\Rightarrow x=y=z=\dfrac{\sqrt{2}}{3}\)
`P=1/(x^2+y^2)+1/(xy)+4xy`
`=1/(x^2+y^2)+1/(2xy)+4xy+1/(4xy)+1/(4xy)`
Áp dụng bunhia dạng phân thức
`=>1/(x^2+y^2)+1/(2xy)>=4/(x+y)^2`
Mà `(x+y)^2<=1`
`=>1/(x^2+y^2)+1/(2xy)>=4`
Áp dụng cosi:
`4xy+1/(4xy)>=2`
`4xy<=(x+y)^2<=1`
`=>1/(4xy)>=1`
`=>P>=4+2+1=7`
Dấu "=" `<=>x=y=1/2`
Ta có \(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left(\dfrac{\left(a+b\right)^2}{2}\right)^2}{2}=\dfrac{\left(a+b\right)^4}{8}\). Áp dụng cho biểu thức A, suy ra \(A\ge\dfrac{\left(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\right)^4}{8}\). Ta tìm GTNN của \(P=x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\). Ta có
\(P=x^2+\dfrac{1}{16x^2}+y^2+\dfrac{1}{16y^2}+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+2\)
\(P\ge2\sqrt{x^2.\dfrac{1}{16x^2}}+2\sqrt{y^2.\dfrac{1}{16y^2}}+\dfrac{15}{16}\left(\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2}{2}\right)+2\)
\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{15}{16}.\left(\dfrac{4^2}{2}\right)+2\) \(=\dfrac{21}{2}\). Do đó \(P\ge\dfrac{21}{2}\) \(\Leftrightarrow A\ge\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\). Vậy GTNN của A là \(\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\), ĐTXR \(\Leftrightarrow x=y=\dfrac{1}{2}\)
Gợi ý: \(\dfrac{a^4+b^4}{2}\ge\left(\dfrac{a+b}{2}\right)^4\)
\(A=\dfrac{4}{2-x}+\dfrac{100}{x}+2021=36\left(2-x\right)+\dfrac{4}{2-x}+36x+\dfrac{100}{x}+1949\)
\(0< x< 2\Rightarrow\left\{{}\begin{matrix}x>0\\x< 2\Rightarrow-x>-2\Leftrightarrow2-x>0\end{matrix}\right.\)
\(\Rightarrow A\ge2\sqrt{36\left(2-x\right).\dfrac{4}{\left(2-x\right)}}+2\sqrt{36x.\dfrac{100}{x}}+1985=2\sqrt{4.36}+2\sqrt{36.100}+1949=2093\Rightarrow A_{min}=2093\Leftrightarrow\left\{{}\begin{matrix}36\left(2-x\right)=\dfrac{4}{2-x}\\36x=\dfrac{100}{x}\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{5}{3}\left(tm\right)\)
\(10x^2+\frac{1}{x^2}+\frac{y^2}{4}=20\)
\(=>\left(x^2+\frac{1}{x^2}\right)+\left(9x^2+\frac{y^2}{4}\right)=20\)
\(=>\left(x+\frac{1}{x}\right)^2+\left(3x+\frac{y}{2}\right)^2=20\)
Ta có \(x+\frac{1}{x}\ge2\sqrt{\frac{x.1}{x}}\ge2\)dấu = xảy ra khi x=1
=> y=6
=> MinP=6
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