Question

Cho a,b,c > 0 và a+b+c=1

Tìm  max \(\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{c+a}\)

Answer 1

Ta có:

\(\sqrt[3]{a+b}=\sqrt[3]{\frac{9}{4}}.\sqrt[3]{\left(a+b\right).\frac{2}{3}.\frac{2}{3}}\le\frac{\left(a+b\right)+\frac{2}{3}+\frac{2}{3}}{3}\)

Tương tự:

\(\sqrt[3]{b+c}\le\frac{\left(b+c\right)+\frac{2}{3}+\frac{2}{3}}{3}\)

\(\sqrt[3]{c+a}\le\frac{\left(c+a\right)+\frac{2}{3}+\frac{2}{3}}{3}\)

\(\Rightarrow\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{c+a}\le\sqrt[3]{\frac{9}{4}}.\frac{2\left(a+b+c\right)+4}{3}\)

\(=\sqrt[3]{\frac{9}{4}}.\frac{6}{3}=\sqrt[3]{18}\)

(Dấu "="\(\Leftrightarrow\hept{\begin{cases}a+b=\frac{2}{3}\\b+c=\frac{2}{3}\\c+a=\frac{2}{3}\end{cases}}\)\(\Leftrightarrow a=b=c=\frac{1}{3}\))

Answer 2

Em làm sai tại đây nhé:

\(\sqrt[3]{a+b}=\sqrt[3]{\frac{9}{4}}.\sqrt[3]{\left(a+b\right).\frac{2}{3}.\frac{2}{3}}\le\sqrt[3]{\frac{9}{4}}.\frac{1}{3}.\left(a+b+\frac{2}{3}+\frac{2}{3}\right)\)

Answer 3

Thêm giùm mình \(.\sqrt[3]{\frac{9}{4}}\)ở ba bđt nhé

Như vậy thì sẽ đúng