Áp dụng BĐT Cô-si ta có
\(\dfrac{a}{a^2+1}\) + \(\dfrac{5\left(a^2+1\right)}{2a}\) \(\ge\sqrt{\dfrac{5}{2}}\)
Dấu "=" xảy ra <=> 2a2 = ( a2 +1 )2
=>\(\left[{}\begin{matrix}a^2+1=2a\\a^2+1=-2a\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}a^2-2a+1=0\\a^2+2a+1=0\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}\left(a-1\right)^2=0\\\left(a+1\right)^2=0\end{matrix}\right.\) => (a - 1)2 = 0 (vì a + 1 >0)
=> a = 1
Vậy Pmin = \(\sqrt{\dfrac{5}{2}}\) <=>a = 1
P = \(\dfrac{a}{a^2+1}\) + \(\dfrac{a^2+1}{4a}\) + \(\dfrac{9\left(a^2+1\right)}{4a}\)
Cô-si 2 con đầu ra a = 1
thay a = 1 => P = \(\dfrac{11}{2}\)
