Đề bài: $a/$Cho $\begin{cases}x+y\geq 2 \\ x,y\geq 0 \\n\in N^{*}\end{cases}$Chứng minh: $x^{n+1}+y^{n+1}\geq x^{n}+y^{n}$$b/$Cho $\begin{cases} \\a,b> 0 \\n\in N^{*}\end{cases}$Chứng minh: $\frac{a^{n}+b^{n}}{2}\geq \left ( \frac{a+b}{2} \right )^{n}$

Lời giải

Đề bài:
$a/$Cho $\begin{cases}x+y\geq 2 \\ x,y\geq 0 \\n\in N^{*}\end{cases}$Chứng minh: $x^{n+1}+y^{n+1}\geq x^{n}+y^{n}$$b/$Cho $\begin{cases} \\a,b> 0 \\n\in N^{*}\end{cases}$Chứng minh: $\frac{a^{n}+b^{n}}{2}\geq \left ( \frac{a+b}{2} \right )^{n}$
Lời giải

$a/$Ta có:
$2\left ( x^{n+1}+ y^{n+1}\right )-\left ( x^{n}+ y^{n}\right )\left ( x +y\right )=x^{n}\left ( x -y\right )-y^{n}\left ( x -y\right )$
$=\left ( x^{n}- y^{n}\right )\left ( x -y\right )=\left ( x -y\right )^{2}\left ( x^{n-1}+ x^{n-2}y+…+ y^{n-2}x+ y^{n-1}\right )\geq 0$
$\Rightarrow 2\left ( x^{n+1}+ y^{n+1}\right )\geq \left ( x^{n}+ y^{n}\right )\left ( x +y\right )\geq 2\left ( x^{n}+ y^{n}\right )$
$\Rightarrow x^{n+1}+ y^{n+1}\geq x^{n}+ y^{n}$
$b/$Đặt $\begin{cases}x= \frac{2a}{a+b}>0\\ y= \frac{2b}{a+b}>0 \end{cases}\Rightarrow x+y=2$
Theo câu a:
$x^{n}+ y^{n}\geq x^{n-1}+ y^{n-1}\geq x+y=2$
$\Rightarrow x^{n}+ y^{n}\geq 2\Rightarrow \left (  \frac{2a}{a+b} \right )^{n}+ \left (  \frac{2b}{a+b} \right )^{n}\geq 2\Leftrightarrow \frac{a^{n}+b^{n}}{2}\geq \left ( \frac{a+b}{2} \right )^{n}$

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