Question

Với a+b+c+d = 1. Tìm Max 

A = \(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}+\sqrt{4d+1}\)

Answer 1

Áp dụng BĐT \(ab\le\dfrac{a^2+b^2}{2}\) , ta có:

\(\sqrt{2}.\sqrt{4a+1}\le\dfrac{2+4a+1}{2}=\dfrac{4a+3}{2}\)

Tương tự:

\(\sqrt{2}.\sqrt{4b+1}\le\dfrac{4b+3}{2}\)

\(\sqrt{2}.\sqrt{4c+1}\le\dfrac{4c+3}{2}\)

\(\sqrt{2}.\sqrt{4d+1}\le\dfrac{4d+3}{2}\)

\(\Rightarrow\sqrt{2}\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}+\sqrt{4d+1}\right)\le\dfrac{4\left(a+b+c\right)+12}{2}=\dfrac{16}{2}=8\)

\(\Rightarrow\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}+\sqrt{4d+1}\right)\le4\sqrt{2}\)

\(MaxA=4\sqrt{2}\Leftrightarrow a=b=c=d=\dfrac{1}{4}\)