\(P=\dfrac{1}{a^2+b^2+1}+\dfrac{1}{2ab}=\dfrac{1}{\left(a+b\right)^2-2ab+1}+\dfrac{1}{2ab}\ge\dfrac{1}{2-2ab}+\dfrac{1}{2ab}\)
Áp dụng BĐt cauchy :
\(\dfrac{1}{2-2ab}+\dfrac{1}{2ab}\ge\dfrac{4}{2-2ab+2ab}=2\)
dấu = xảy ra khi \(\left\{{}\begin{matrix}a+b=1\\2ab=2-2ab\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=1\\ab=\dfrac{1}{2}\end{matrix}\right.\)......
AD bất đẳng thức Cauchy Schwarz, ta có
P=\(\dfrac{1}{a^2+b^2+1}+\dfrac{1}{2ab}\)\(\ge\dfrac{\left(1+1\right)^2}{a^2+2ab+b^2+1}\)=\(\dfrac{4}{\left(a+b\right)^2+1}\)(*)
mà a+b\(\le1\)=>(*)\(\ge\)\(\dfrac{4}{1+1}=2\)
