Ta có  : \(\frac{1+x}{2}\ge\sqrt{x}\Rightarrow\left(\frac{1+x}{2}\right)^n\ge\sqrt{x^n}\) (1)

            \(\frac{1+y}{2}\ge\sqrt{y}\Rightarrow\left(\frac{1+y}{2}\right)^n\ge\sqrt{y^n}\)(2)

            \(\frac{1+z}{2}\ge\sqrt{z}\Rightarrow\left(\frac{1+z}{2}\right)^n\ge\sqrt{z^n}\)(3) 

Từ 1,2,3 \(\Rightarrow\left(\frac{1+x}{2}\right)^n+\left(\frac{1+y}{2}\right)^n+\left(\frac{1+z}{2}\right)^n\ge\sqrt{x^n}+\sqrt{y^n}+\sqrt{z^n}\)

Áp dụng BĐT Cauchy cho 3 số ta có : 

\(\sqrt{x^n}+\sqrt{y^n}+\sqrt{z^n}\ge3^3\sqrt{\sqrt{x^n}.\sqrt{y^n}.\sqrt{z^n}}=3\)

\(\Rightarrow\left(\frac{1+x}{2}\right)^n+\left(\frac{1+y}{2}\right)^n+\left(\frac{1+z}{2}\right)^n\ge3\)

Đẳng thức xảy ra <=> x = y = z = 1 

Bài cuối có Max nữa nhé, cần thì ib mình làm cho.

Giả sử \(c=min\left\{a;b;c\right\}\Rightarrow c\le1< 2\Rightarrow2-c>0\)

Ta có:\(P=ab+bc+ca-\frac{1}{2}abc=\frac{ab}{2}\left(2-c\right)+bc+ca\ge0\)

Đẳng thức xảy ra tại \(a=3;b=0;c=0\) và các hoán vị

3/ \(P=\Sigma\frac{\left(3-a-b\right)\left(a-b\right)^2}{3}+\frac{5}{2}abc\ge0\)

Bài 1 : 

Ta có :

\(\left(x-1\right)^6=\left(x-1\right)^8\)

\(\Leftrightarrow\)\(x-1=\left(x-1\right)^2\)

\(\Leftrightarrow\)\(\left(x-1\right)-\left(x-1\right)^2=0\)

\(\Leftrightarrow\)\(\left(x-1\right)\left(1-x+1\right)=0\)

\(\Leftrightarrow\)\(\left(x-1\right)\left(2-x\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-1=0\\2-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)

Vậy \(x=1\) hoặc \(x=2\)

a) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)

\(\Rightarrow\left(\frac{1}{3}\right)^n=\frac{1^4}{3^4}\)

\(\Rightarrow\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^4\)

\(\Rightarrow n=4\)

Vậy n = 4

b) \(\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)

\(\Rightarrow\frac{-8^3}{7^3}=\left(\frac{-8}{7}\right)^n\)

\(\Rightarrow\left(\frac{-8}{7}\right)^3=\left(\frac{-8}{7}\right)^n\)

\(\Rightarrow n=3\)

Vậy n = 3