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Ta có:
Đặt \(A=x+y+\dfrac{1}{x}+\dfrac{1}{y}\)
\(\Leftrightarrow A=x+y+\dfrac{4}{4x}+\dfrac{4}{4y}\)
\(\Leftrightarrow A=x+y+\dfrac{1}{4x}+\dfrac{3}{4x}+\dfrac{1}{4y}+\dfrac{3}{4y}\)
\(\Leftrightarrow A=\left(x+\dfrac{1}{4x}\right)+\left(y+\dfrac{1}{4y}\right)+\left(\dfrac{3}{4x}+\dfrac{3}{4y}\right)\)
\(\Rightarrow A\ge2\sqrt{x.\dfrac{1}{4x}}+2\sqrt{y.\dfrac{1}{4y}}+\dfrac{3}{4}.\dfrac{4}{x+y}\)
\(\ge2.\sqrt{\dfrac{1}{4}}+2\sqrt{\dfrac{1}{4}}+\dfrac{3}{4}.\dfrac{4}{1}\)
\(=2.\dfrac{1}{2}+2.\dfrac{1}{2}+3=1+1+3=5\)
Vậy ta có đpcm. Dấu"=" xảy ra\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4x}\\y=\dfrac{1}{4y}\\x=y\\x+y=1\end{matrix}\right.\) \(\Leftrightarrow x=y=\dfrac{1}{2}\left(tm\right)\)
Đặt \(A=x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)
\(\Leftrightarrow A=x+y+z+\dfrac{9}{9x}+\dfrac{9}{9y}+\dfrac{9}{9z}\)
\(\Leftrightarrow A=x+y+z+\dfrac{1}{9x}+\dfrac{8}{9x}+\dfrac{1}{9y}+\dfrac{8}{9y}+\dfrac{1}{9z}+\dfrac{8}{9z}\)
\(\Leftrightarrow A=\left(x+\dfrac{1}{9x}\right)+\left(y+\dfrac{1}{9y}\right)+\left(z+\dfrac{1}{9z}\right)+\left(\dfrac{8}{9x}+\dfrac{8}{9y}+\dfrac{8}{9z}\right)\)
\(\Leftrightarrow A=\left(x+\dfrac{1}{9x}\right)+\left(y+\dfrac{1}{9y}\right)+\left(z+\dfrac{1}{9z}\right)+\dfrac{8}{9}.\left(\dfrac{1^2}{x}+\dfrac{1^2}{y}+\dfrac{1^2}{z}\right)\)
\(\Rightarrow A\ge2\sqrt{x.\dfrac{1}{9x}}+2\sqrt{y.\dfrac{1}{9y}}+2\sqrt{z.\dfrac{1}{9z}}+\dfrac{8}{9}.\dfrac{\left(1+1+1\right)^2}{x+y+z}\)
\(\Rightarrow A\ge2\sqrt{\dfrac{1}{9}}+2\sqrt{\dfrac{1}{9}}+2\sqrt{\dfrac{1}{9}}+\dfrac{8}{9}.\dfrac{3^2}{1}\)
\(\Rightarrow A\ge2.\dfrac{1}{3}.3+8=2+8=10\)
Vậy ta có BĐT cần chứng minh.
Dấu\("="\) xảy ra\(\Leftrightarrow x=y=z=\dfrac{1}{3}\)
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\(P=\sqrt{x^4+x^2y^2}+x^2=\sqrt{x^4+\frac{1}{x^2}}+x^2\)
Ta có: \(x^4+\frac{1}{x^2}=x^4+\frac{1}{8x^2}+\frac{1}{8x^2}+...+\frac{1}{8x^2}\ge9\sqrt[9]{x^4.\left(\frac{1}{8x^2}\right)^8}\)
\(=9\sqrt[9]{\frac{1}{8^8.x^{12}}}\)
=> \(P=3\sqrt[18]{\frac{1}{8^8.x^{12}}}+x^2\)
\(=\sqrt[18]{\frac{1}{8^8x^{12}}}+\sqrt[18]{\frac{1}{8^8x^{12}}}+\sqrt[18]{\frac{1}{8^8x^{12}}}+x^2\)
\(\ge4\sqrt[4]{\left(\sqrt[18]{\frac{1}{8^8x^{12}}}\right)^3.x^2}\)
\(=4.\left(\frac{1}{8^{\frac{1}{3}}.x^{\frac{1}{2}}}\right).x^2=2\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x^4=\frac{1}{8x^2}\\x^2=\sqrt[8]{\frac{1}{8^8x^{12}}}\end{cases}}\)<=> x^2 = 1/2 khi đó y = 2 , x = \(\frac{1}{\sqrt{2}}\)
Vậy GTNN của P = 2.
Lời giải:
$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=2$
$\Leftrightarrow (x+\sqrt{x^2+1})(x-\sqrt{x^2+1})(y+\sqrt{y^2+1})=2(x-\sqrt{x^2+1})$
$\Leftrightarrow -(y+\sqrt{y^2+1})=2(x-\sqrt{x^2+1})$
$\Leftrightarrow 2x+\sqrt{y^2+1}=2\sqrt{x^2+1}-y$
$\Rightarrow (2x+\sqrt{y^2+1})^2=(2\sqrt{x^2+1}-y)^2$
$\Leftrightarrow 4x^2+y^2+1+4x\sqrt{y^2+1}=4(x^2+1)+y^2-4y\sqrt{x^2+1}$
$\Leftrightarrow 4(x\sqrt{y^2+1})+y\sqrt{x^2+1})=3$
$\Leftrightarrow 4Q=3$
$\Leftrightarrow Q=\frac{3}{4}$
Đáp án đúng : B