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 \(\sqrt{x^2+1}=\dfrac{1}{\sqrt{x^2+1}}\Rightarrow x=0\)
\(b,\dfrac{2x^2+1}{\sqrt{4x^2+1}}\ge\dfrac{2x^2+1}{\dfrac{4x^2+1+1}{2}}=\dfrac{2x^2+1}{2x^2+1}=1\)
Vậy \(\dfrac{2x^2+1}{\sqrt{4x^2+2}}\ge1\)
Phần b mình làm linh tinh thôi
