Vương Ngọc Như Quỳnh
29/08/2017 at 20:53-
VTK-VangTrangKhuyet 29/08/2017 at 20:56Using Bunhiakovski's inequaility :
\(\left(1\cdot\sqrt{4a+1}+1\cdot\sqrt{4b+1}\right)^2\le\left(1^2+1^2\right)\cdot\left(4a+1+4b+1\right)=2\cdot6=12\)
=> \(\sqrt{4a+1}+\sqrt{4b+1}\le\sqrt{12}\)
The "=" happens when a = b = \(\dfrac{1}{2}\)
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