chứng minh $\dfrac{x^n +y^n +z^n}{3}\ge (\dfrac{x+y+z}{3})^n$

S

sieumau88

Dùng quy nạp để c/m $\dfrac{x^n +y^n +z^n}{3}$ \geq $\left(\dfrac{x+y+z}{3}\right)^n$ (*)

Biết rằng $x,y,n$ là các số nguyên dương và $n \in N$ .
________________________

:cool: Với $n=1$ ta có (*) đúng .

:cool: Ta sẽ c/m (*) cũng đúng với $n=k+1$ , nghĩa là $\dfrac{{x}^{n+1} + {y}^{n+1} + {z}^{n+1}}{3}$ \geq ${\left(\dfrac{x+y+z}{3}\right)}^{n+1}$ (**)

Nhân 2 vế của (*) cho $(x+y+z)$ ta có $\dfrac{(x^n +y^n +z^n).(x+y+z)}{3}$ \geq $\dfrac{{\left({x+y+z}\right)}^{n+1}}{3^n}$

\Leftrightarrow $\dfrac{{x}^{n+1}+{y}^{n+1}+{z}^{n+1}+xy^n + xz^n+yx^n+yz^n+zx^n+zy^n}{3}$ \geq $\dfrac{{\left({x+y+z}\right)}^{n+1}}{3^n}$

\Leftrightarrow ${x}^{n+1}+{y}^{n+1}+{z}^{n+1}$ \geq $\dfrac{{\left({x+y+z}\right)}^{n+1}}{3^n}$

chia 2 vế cho 3 \Leftrightarrow $\dfrac{{x}^{n+1} + {y}^{n+1} + {z}^{n+1}}{3}$ \geq ${\left(\dfrac{x+y+z}{3}\right)}^{n+1}$
 
Last edited by a moderator:
C

chonhoi110

Cách khác :
Áp dụng bđt $AM-GM$ ta có:
$x^n+(n-1)(\dfrac{x+y+z}{3})^n\ge n. \sqrt[n]{x^n(\dfrac{x+y+z}{3}) ^{n(n-1)}}= n(\dfrac{x+y+z}{3})^{n-1}x$

$y^n+(n-1)(\dfrac{x+y+z}{3})^n\ge n. \sqrt[n]{y^n(\dfrac{x+y+z}{3}) ^{n(n-1)}}= n(\dfrac{x+y+z}{3})^{n-1}y$

$z^n+(n-1)(\dfrac{x+y+z}{3})^n\ge n. \sqrt[n]{z^n(\dfrac{x+y+z}{3}) ^{n(n-1)}}= n(\dfrac{x+y+z}{3})^{n-1}z$

$\rightarrow (x^n+y^n+z^n) \ge n(\dfrac{x+y+z}{3})^{n-1} (x+y+z)-3(n-1)(\dfrac{x+y+z}{3})^n=3(\dfrac{x+y+z}{3})^n$

P/s: Trích trong sách Những viên kim cương trong bđt @.@
 
You must log in or register to reply here.
Top Bottom