Cô si thôi:

\(0\le\left(b+c-a\right)\left(c+a-b\right)\le\frac{\left(b+c-a\right)+\left(c+a-b\right)}{2}=c\)

\(0\le\left(c+a-b\right)\left(a+b-c\right)\le\frac{\left(c+a-b\right)+\left(a+b-c\right)}{2}=a\)

\(0\le\left(b+c-a\right)\left(a+b-c\right)\le\frac{\left(b+c-a\right)+\left(a+b-c\right)}{2}=b\)

\(\Rightarrow0\le\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)\le abc\)

(Dấu "=" khi và chỉ khi a = b = c hay tam giác ABC đều)

<=>3\(\sqrt{2x}\)-20\(\sqrt{2x}\)+21\(\sqrt{2x}\)=28

<=>4\(\sqrt{2x}\)=28

<=>\(\sqrt{2x}\)=7

<=>2x=14

<=>x=7

\(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28\)

\(3\sqrt{2x}-5\sqrt{8}.\sqrt{x}+7\sqrt{18x}=28\)

\(3\sqrt{2x}-5.2\sqrt{2}.\sqrt{x}+7\sqrt{18x}=28\)

\(3\sqrt{2x}-5.2\sqrt{2}.\sqrt{x}+7.\sqrt{18}.\sqrt{x}=28\)

\(3\sqrt{2x}-5.2\sqrt{2}.\sqrt{x}+7.3\sqrt{2}.\sqrt{x}=28\)

\(3\sqrt{2x}-5.2\sqrt{2x}+7.3\sqrt{2x}=28\)

\(3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)

\(14\sqrt{2x}=28\)

\(392x=784\)

\(x=\frac{784}{392}=2\)

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

Em làm sai tại đây nhé:

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

câu c ghi có đúng ko vậy bạn

đúng rồi bạn