Question

Answer 1

\(Áp\ dụng\ BĐT\ Cô\ Si\ cho\ 2\ số\ thực\ dương,\ ta\ được :\)

\(a^2+b^2\ge2\sqrt{a^2.b^2}=2ab\\ < =>4\ge2ab\\ < =>ab\le2\)

\(Ta\ có :\)

\(a^2+b^2=4\\ < =>\left(a+b\right)^2=4+2ab\le4+2.2=8\\ < =>0< a+b\le2\sqrt{2}\)

\(Lại\ có:\)

 \(a^2+b^2=4\\ < =>\left(a+b\right)^2=4+2ab\\ < =>2ab=\left(a+b\right)^2-4\\ < =>ab=\dfrac{\left(a+b-2\right)\left(a+b+2\right)}{2}\)

\(=>T=\dfrac{\dfrac{\left(a+b-2\right)\left(a+b+2\right)}{2}}{a+b+2}=\dfrac{a+b-2}{2}\le\dfrac{2\sqrt{2}-2}{2}=\sqrt{2}-1\)

\(Dấu ''=''\ xảy\ ra\ khi :\) \(\left\{{}\begin{matrix}a^2=b^2\\a^2+b^2=4\\a,b>0\end{matrix}\right.< =>a=b=\sqrt{2}\)

\(Vậy\ GTLN\ của\ T\ là: \) \(\sqrt{2}-1< =>a=b=\sqrt{2}\)