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Áp dụng BĐT AM-GM, Ta có
\(\sqrt{x-1}\le\dfrac{1+x-1}{2}=\dfrac{x}{2}\Rightarrow yz\sqrt{x-1}\le\dfrac{xyz}{2}\)
Mà \(xz\sqrt{y-2}\le\dfrac{xz\sqrt{2\left(y-2\right)}}{\sqrt{2}}\le\dfrac{xyz}{2\sqrt{2}}\)
\(yx\sqrt{z-3}\le yx.\dfrac{3+z-3}{2\sqrt{3}}=\dfrac{xyz}{2\sqrt{3}}\)
\(\Rightarrow\dfrac{xy\sqrt{x-1}+xz\sqrt{y-2}+yz\sqrt{z-3}}{xyz}\le\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}=\dfrac{1}{2}+\dfrac{\sqrt{2}}{4}+\dfrac{\sqrt{3}}{6}\)
Đề bài sai: Khi \(x=4\) thì \(A=\dfrac{1}{2};B=\dfrac{28}{9};\dfrac{A}{B}=\dfrac{9}{56};\dfrac{x-2}{4\sqrt{x}}=\dfrac{1}{4}\Rightarrow\dfrac{A}{B}\ne\dfrac{x-2}{4\sqrt{x}}\)
c.
\(\left(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right)^2=2010\)
\(\leftrightarrow\) \(x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+1+x^2+y^2+x^2y^2=2010\)
\(\leftrightarrow\)\(x^2+x^2y^2+2x\sqrt{1+y^2}.y\sqrt{1+x^2}+y^2+x^2y^2=2009\)
\(\leftrightarrow\) \(\left(x\sqrt{1+y^2}+y\sqrt{1+x^2}\right)^2=2009\)
\(\leftrightarrow\) \(x\sqrt{1+y^2}+y\sqrt{1+x^2}=\sqrt{2009}\)
c) \(A^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=x^2y^2+x^2+x^2y^2+y^2+1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-1\)
\(=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-1\)
\(=\left[xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right]^2-1=2010-1=2009\)
Vì A>0 nên \(A=\sqrt{2009}\)
d) \(2009^2=\left(2008+1\right)^2=2008^2+2.2008+1\)
\(1+2008^2=2009^2-2.2008=2009^2-2.2009\dfrac{2008}{2009}\)
\(A=\sqrt{2009^2-2.2009.\dfrac{2008}{2009}+\dfrac{2008^2}{2009^2}}+\dfrac{2008}{2009}\)
\(A=\sqrt{\left(2009-\dfrac{2008}{2009}\right)^2}+\dfrac{2008}{2009}=2009-\dfrac{2008}{2009}+\dfrac{2008}{2009}=2009\)
a) \(\dfrac{x^2+2}{\sqrt{x^2+1}}\ge2\) \(\Leftrightarrow\) \(x^2+2\ge2\sqrt{x^2+1}\)
\(\Leftrightarrow\) \(\left(x^2+2\right)^2\ge\left(2\sqrt{x^2+1}\right)^2\) \(\Leftrightarrow\) \(x^4+4x^2+4\ge4x^2+4\)
\(\Leftrightarrow\) \(x^4\ge0\) (đúng \(\forall x\)) \(\Rightarrow\) \(\dfrac{x^2+2}{\sqrt{x^2+1}}\ge2\) (đpcm)
b) \(\dfrac{2x^2+1}{\sqrt{4x^2+1}}\ge1\) \(\Leftrightarrow\) \(2x^2+1\ge\sqrt{4x^2+1}\)
\(\Leftrightarrow\) \(\left(2x^2+1\right)^2\ge\left(\sqrt{4x^2+1}\right)^2\) \(\Leftrightarrow\) \(4x^4+4x^2+1\ge4x^2+1\)
\(\Leftrightarrow\) \(4x^4\ge0\) (đúng \(\forall x\)) \(\Rightarrow\) \(\dfrac{2x^2+1}{\sqrt{4x^2+1}}\ge1\) (đpcm)
a,
\(\dfrac{x^2+2}{\sqrt{x^2+1}}=\dfrac{\left(\sqrt{x^2+1}\right)^2+1}{\sqrt{x^2+1}}=\sqrt{x^2+1}+\dfrac{1}{\sqrt{x^2+1}}\ge2\)( Áp dụng bất đẳng thức AM - GM )
Vậy:
\(\dfrac{x^2+1}{\sqrt{x+1}}\ge2\)
Đẳng thức xảy ra khi và chỉ khi \(\sqrt{x^2+1}=\dfrac{1}{\sqrt{x^2+1}}\Rightarrow x=0\)