chứng minh x>0 thì (x+1)^2 (1/x^2 +2/x+1)>=16 02/08/2021 Bởi Lydia chứng minh x>0 thì (x+1)^2 (1/x^2 +2/x+1)>=16
Giải thích các bước giải: Ta có:$(x+1)^2(\dfrac{1}{x^2}+\dfrac{2}{x}+1)=(x+1)^2(\dfrac1x+1)^2$Lại có:$\dfrac1x+1=\dfrac1x+\dfrac11\ge \dfrac4{x+1}$$\to (x+1)(\dfrac1x+1)\ge 4$$\to ((x+1)(\dfrac1x+1)^2\ge 16$$\to (x+1)^2(\dfrac1x+1)^2\ge 16$$\to (x+1)^2(\dfrac{1}{x^2}+\dfrac2x+1)\ge 16$ Bình luận
Giải thích các bước giải:
Ta có:
$(x+1)^2(\dfrac{1}{x^2}+\dfrac{2}{x}+1)=(x+1)^2(\dfrac1x+1)^2$
Lại có:
$\dfrac1x+1=\dfrac1x+\dfrac11\ge \dfrac4{x+1}$
$\to (x+1)(\dfrac1x+1)\ge 4$
$\to ((x+1)(\dfrac1x+1)^2\ge 16$
$\to (x+1)^2(\dfrac1x+1)^2\ge 16$
$\to (x+1)^2(\dfrac{1}{x^2}+\dfrac2x+1)\ge 16$