Question

Bài 1 Cho a + b+ c = 3 ; a,b,c > 0

Tìm GTLN của

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

áp dụng bđt Cô-Si nhé

 

 

Answer 1

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

\(=\dfrac{\sqrt{2\left(a+b\right)}}{\sqrt{2}}+\dfrac{\sqrt{2\left(b+c\right)}}{\sqrt{2}}+\dfrac{\sqrt{2\left(c+a\right)}}{\sqrt{2}}\)

\(\le\dfrac{\dfrac{2+\left(a+b\right)}{2}}{\sqrt{2}}+\dfrac{\dfrac{2+\left(b+c\right)}{2}}{\sqrt{2}}+\dfrac{\dfrac{2+\left(c+a\right)}{2}}{\sqrt{2}}\)

\(=\dfrac{6+2\left(a+b+c\right)}{2\sqrt{2}}=\dfrac{6+2.3}{2\sqrt{2}}=3\sqrt{2}\)

- Vậy \(MaxQ=3\sqrt{2}\), đạt tại \(a=b=c=1\)