\(1,yz\sqrt{x-1}=yz\sqrt{\left(x-1\right)\cdot1}\le yz\cdot\dfrac{x-1+1}{2}=\dfrac{xyz}{2}\)

\(zx\sqrt{y-2}=\dfrac{zx\cdot2\sqrt{2\left(y-2\right)}}{2\sqrt{2}}\le\dfrac{xyz}{2\sqrt{2}}\\ xy\sqrt{z-3}=\dfrac{xy\cdot2\sqrt{3\left(z-3\right)}}{2\sqrt{3}}\le\dfrac{xyz}{2\sqrt{3}}\)

\(\Leftrightarrow M\le\dfrac{\dfrac{xyz}{2}+\dfrac{xyz}{2\sqrt{2}}+\dfrac{xyz}{2\sqrt{3}}}{xyz}=\dfrac{xyz\left(\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\right)}{xyz}=\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=2\\z-3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\\z=6\end{matrix}\right.\)

\(2,N^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\\ \Leftrightarrow N^2\le\left(a+b+b+c+c+a\right)\left(1^2+1^2+1^2\right)\\ \Leftrightarrow N^2\le6\left(a+b+c\right)=6\sqrt{2}\\ \Leftrightarrow N\le\sqrt{6\sqrt{2}}\)

Dấu \("="\Leftrightarrow a=b=c=\dfrac{\sqrt{2}}{3}\)

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

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

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

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

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:

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

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

Như vậy thì sẽ đúng

mong mn giúp mk vs 

Có \(\sqrt{2a+b+1}\le\frac{2a+b+1+4}{4}\)
Tương tự \(\sqrt{2b+c+1}\le\frac{2b+c+1+4}{4},\sqrt{2c+a+1}\le\frac{2c+a+1+4}{4}\)
\(\Rightarrow A\le\frac{2a+b+1+2c+a+1+2b+c+1+4+4+4}{4}=6\)
dấu = xảy ra khi a=b=c và a+b+c=3=>a=b=c=1

Áp dụng BĐT Cauchy-Schwarz ta có:

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

\(\le\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(1+1+1\right)\)

\(=3\cdot2\left(a+b+c\right)=6\cdot4=24\)

\(\Rightarrow A^2\le24\Rightarrow A\le\sqrt{24}\)

Đẳng thức xảy ra khi \(a=b=c=\dfrac{4}{3}\)

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- Max

Áp dụng bất đẳng thức Cauchy Shwarz, ta có:

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

\(\Rightarrow A^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\)

\(\le\left(1+1+1\right)\left(a+b+b+c+c+a\right)=6\)

\(\Rightarrow A\le\sqrt{6}\left(A>0\right)\)

Mới tối hôm qua làm bên AoPS-_- Sửa đề: a, b, c \(\ge\)0.

Bài làm:(bên đó tên níc của em là SBM):inequality!

Đăng ảnh lên cho dễ xem nha!(ko chắc lắm đâu, đây là phần min)