Question

cho a,b>0 và a+b<=1, tìm GTNN của P=1/(a^2+b^2+1)+1/2ab

Answer 1

\(P=\dfrac{1}{a^2+b^2+1}+\dfrac{1}{2ab}=\dfrac{1}{\left(a+b\right)^2-2ab+1}+\dfrac{1}{2ab}\ge\dfrac{1}{2-2ab}+\dfrac{1}{2ab}\)

Áp dụng BĐt cauchy :

\(\dfrac{1}{2-2ab}+\dfrac{1}{2ab}\ge\dfrac{4}{2-2ab+2ab}=2\)

dấu = xảy ra khi \(\left\{{}\begin{matrix}a+b=1\\2ab=2-2ab\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=1\\ab=\dfrac{1}{2}\end{matrix}\right.\)......

Answer 2

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\(\le1\)=>(*)\(\ge\)\(\dfrac{4}{1+1}=2\)