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

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

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

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

mong mn giúp mk vs 

\(a+b+c+2=abc\)

\(\Leftrightarrow2a+2b+2c+3+ab+bc+ca=abc+ab+bc+ca+a+b+c+1\)

\(\Leftrightarrow\left(a+1\right)\left(b+1\right)+\left(c+1\right)\left(b+1\right)+\left(c+1\right)\left(a+1\right)=\left(a+1\right)\left(b+1\right)\left(c+1\right)\)

\(\Leftrightarrow\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}=1\)

Đặt \(\left(\dfrac{1}{a+1};\dfrac{1}{b+1};\dfrac{1}{c+1}\right)=\left(x;y;z\right)\)

\(\Rightarrow x+y+z=1\)

BĐT trở thành:

\(P=\sqrt{x}+\sqrt{y}+\sqrt{z}\le\sqrt{3\left(x+y+z\right)}=\sqrt{3}\)

Dấu "=" xảy ra khi và chỉ khi \(x=y=z=\dfrac{1}{3}\) hay \(a=b=c=2\)

Ta có \(4\sqrt[3]{a+7}\le\frac{a+7+8+8}{3}=\frac{a+23}{3}\)

\(4\sqrt[3]{b+7}\le\frac{b+23}{3}\)

Từ đó ta có

\(4P=4\sqrt[3]{a+7}+4\sqrt[3]{b+7}+4\sqrt[3]{b+7}\)

\(\le\frac{a+b+b+23×3}{3}=\frac{a+2b+23×3}{3}\le24\)

\(\Rightarrow P\le6\)

Đạt được khi a = b = 1

Bạn​ cụ​ thể​ ra đc k

Áp dụng bất đẳng thức cosi cho 3 số không âm là a + 7, 8, 8 ta có

\(a+7+8+8\ge3\sqrt[3]{\left(a+7\right)×8×8}=3×4\sqrt[3]{a+7}\)

\(\Leftrightarrow\frac{a+23}{3}\ge4\sqrt[3]{a+7}\)

Vầy dễ hiểu chưa bạn

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