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Cần điều kiện x;y dương
\(M=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\ge\frac{1}{2}\left(x+\frac{1}{x}+y+\frac{1}{y}\right)^2\)
\(M\ge\frac{1}{2}\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2\ge\frac{1}{2}\left(x+y+\frac{4}{x+y}\right)^2=\frac{25}{2}\)
\(M_{min}=\frac{25}{2}\) khi \(x=y=\frac{1}{2}\)
CHo x,y > 0, x + y = 1. Tìm GTNN của \(S=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)
\(S\ge\frac{1}{2}\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2\ge\frac{1}{2}\left(x+y+\frac{4}{x+y}\right)^2=\frac{1}{2}\left(1+4\right)^2=\frac{25}{2}\)
\(S_{min}=\frac{25}{2}\) khi \(x=y=\frac{1}{2}\)
\(A\)xác định \(\Leftrightarrow x^2y^2+1+\left(x^2-y\right)\left(1-y\right)\ne0\)
\(\Leftrightarrow x^2y^2+1+x^2-x^2y-y+y^2\ne0\)
\(\Leftrightarrow\left(x^2y^2+y^2\right)+\left(x^2+1\right)-\left(x^2y+y\right)\ne0\)
\(\Leftrightarrow y^2\left(x^2+1\right)+\left(x^2+1\right)-y\left(x^2+1\right)\ne0\)
\(\Leftrightarrow\left(x^2+1\right)\left(y^2-y+1\right)\ne0\)
\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\)
Ta có: \(\hept{\begin{cases}x^2+1>0\forall x\\\left(y-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall y\end{cases}}\)\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]>0\forall x;y\)
\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\forall x;y\)
\(\Leftrightarrow A\ne0\forall x;y\)
theo nghiệm Fx=Gx mũ 2
suy ra x mũ 2 +1 mũ x 2
suy ra chịch chịch chịch
Mình ko chắc lắm :
Áp dụng BĐT AM - GM ta có :
\(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)\)
\(=\frac{x^2y^2+1}{y^2}.\frac{x^2y^2+1}{x^2}=\frac{x^4y^4+2x^2y^2+1}{x^2y^2}\)
\(=x^2y^2+\frac{1}{x^2y^2}+2=x^2y^2+\frac{1}{256x^2y^2}+\frac{255}{256x^2y^2}+2\)
\(\ge2\sqrt{x^2y^2.\frac{1}{256x^2y^2}}+\frac{255}{256.\left(xy\right)^2}+2\)
\(\ge2.\frac{1}{16}+\frac{255}{256.\left(\frac{\left(x+y\right)^2}{4}\right)^2}+2\)
\(=\frac{1}{8}+\frac{255}{256.\left(\frac{1}{4}\right)^2}+2=\frac{289}{16}\)
Khi \(x=y=\frac{1}{2}\)
Chúc bạn học tốt !!!
Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :
\(A=\left(1+\frac{1}{x}\right)^2+\left(1+\frac{1}{y}\right)^2\ge\frac{\left(1+\frac{1}{x}+1+\frac{1}{y}\right)^2}{2}=\frac{\left(2+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)(1)
Lại có \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}=\frac{4}{1}=4\)(2)
Từ (1) và (2) => \(A=\left(1+\frac{1}{x}\right)^2+\left(1+\frac{1}{y}\right)^2\ge\frac{\left(2+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\ge\frac{\left(2+4\right)^2}{2}=18\)
Đẳng thức xảy ra <=> x = y = 1/2
Vậy MinA = 18
A = \(\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)
\(=\frac{\left(x+\frac{1}{x}+y+\frac{1}{y}\right)^2}{2}=\frac{1}{2}\left[\left(x+y\right)+\left(\frac{1}{x}+\frac{1}{y}\right)\right]^2\)
\(\ge\frac{1}{2}\left[\left(x+y\right)+\frac{4}{x+y}\right]^2=\frac{1}{2}\left(1+4\right)^2=\frac{25}{2}\)
Dấu "=" xảy ra <=> x = y =1/2
Vậy GTNN của A = 25/2 tại x = y = 1/2
Ta có :
\(A=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)
\(=x^2+\frac{1}{x^2}+2+y^2+\frac{1}{y^2}+2\)
\(=4+\left(x^2+y^2\right)+\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\)
\(\ge4+\frac{\left(x+y\right)^2}{2}+2\sqrt{\frac{1}{\left(xy\right)^2}}\)
\(=4+\frac{1}{2}+\frac{2}{xy}\ge4+\frac{1}{2}+\frac{2}{\frac{\left(x+y\right)^2}{4}}=4+\frac{1}{2}+8=\frac{25}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)
Vậy \(A_{min}=\frac{25}{2}\) tại \(x=y=\frac{1}{2}\)
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