Áp dụng BĐT $Cauchy-schwars$
`P=1/(a^2+b^2)+1/(ab)+4ab`
`P=[1/(a^2+b^2)+1/(2ab)]+[1/(4ab)+4ab]+1/(4ab)`
`P\ge (1+1)^2/(a^2+b^2+2ab)+2\sqrt[1/(4ab). 4ab]+1/(4ab)`
`P\ge 4/(a+b)^2+2\sqrt1+(a+b)^2/(4ab)`
`P\ge 4+2+1=7`
Dấu `=` xảy ra $⇔\begin{cases}a+b=1\\\dfrac{1}{a^2+b^2}=\dfrac{1}{2ab}\\\dfrac{1}{4ab}=4ab\\a=b\end{cases}⇔a=b=\dfrac{1}{2}$
Vậy $Min_P=7⇔a=b=\dfrac{1}{2}$
