5) \(\dfrac{a+8}{\sqrt{a-1}}\ge 6\) \((a>1)\)
Theo bất đẳng thức Cosi: \(a+8=a-1+9\ge2\sqrt{(a-1).9}=6\sqrt{a-1}\)\((a>1)\)
\(\Rightarrow \dfrac{a+8}{\sqrt{a-1}}\ge 6\) \((a>1)\).
6) \(\dfrac{a^2+2}{\sqrt{a^2+1}}\) \(\forall a\)
Theo bất đẳng thức Cosi: \(a^2+2=a^2+1+1\ge2\sqrt{(a^2+1).1}=2\sqrt{a^2+1}\)
\(\Rightarrow \dfrac{a^2+2}{\sqrt{a^2+1}}\).
2) \(\dfrac{bc}{a}+\dfrac{ab}{c}\ge2\sqrt{\dfrac{bc}{a}\dfrac{ab}{c}}=2b\) \((\text{do }b>0\))
\(\dfrac{bc}{a}+\dfrac{ac}{b}\ge2\sqrt{\dfrac{bc}{a}\dfrac{ac}{b}}=2c\) \((\text{do }c>0\))
\(\dfrac{ab}{c}+\dfrac{ac}{b}\ge2\sqrt{\dfrac{ab}{c}\dfrac{ac}{b}}=2a\) \((\text{do }a>0\))
Cộng vế với vế suy ra đpcm.
3) Do \(a+b\ge2\sqrt{ab}\)
\(1+ab\ge2\sqrt{ab.1}=2\sqrt{ab}\)
Với \(a,b\ge0\) nhân vế với vế suy ra \((a+b)(1+ab)\ge4ab\)
