Question

Cho a,b,c >0 tm a+b+c=1.Tìm max \(S=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)

Answer 1

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

\(\text{Tương tự :}\sqrt{b+c}\le\dfrac{\sqrt{3}}{\sqrt{2}}\dfrac{\dfrac{2}{3}+b+c}{2};\sqrt{c+a}\le\dfrac{\sqrt{3}}{\sqrt{2}}\dfrac{\dfrac{2}{3}+c+a}{2}\)

\(\text{Khi đó :}S\le\dfrac{\sqrt{3}}{\sqrt{2}}.\dfrac{2+2\left(a+b+c\right)}{2}=\sqrt{6}\)

\(\text{Vậy maxS=}\sqrt{6}\text{ khi }a=b=c=\dfrac{1}{3}\)