Question

cho a, b, c \(\ge\) 0 và a+b+c = 2022. Tìm Min, Max của:

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

Answer 1

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

\(=\dfrac{1}{\sqrt{1348}}.\sqrt{\left(a+b\right).1348}+\dfrac{1}{\sqrt{1348}}.\sqrt{\left(b+c\right).1348}+\dfrac{1}{\sqrt{1348}}.\sqrt{\left(c+a\right).1348}\)

\(\le\dfrac{1}{\sqrt{1348}}\left[\dfrac{\left(a+b\right)+1348}{2}+\dfrac{\left(b+c\right)+1348}{2}+\dfrac{\left(c+a\right)+1348}{2}\right]\)

\(=\dfrac{1}{\sqrt{1348}}.\left[\left(a+b+c\right)+2022\right]\)

\(=\dfrac{1}{\sqrt{1348}}.\left(2022+2022\right)\)

\(=\dfrac{4044}{\sqrt{1348}}=6\sqrt{337}\)

- Dấu "=" xảy ra khi \(a=b=c=674\)

- Vậy \(MaxA=6\sqrt{337}\)