Chứng minh: `a+(27)/(2(a-1)(a+1)^3)ge5/2( ∀a>1)` 16/11/2021 Bởi Adalyn Chứng minh: `a+(27)/(2(a-1)(a+1)^3)ge5/2( ∀a>1)`
Giải thích các bước giải: Ta có: $A=a+\dfrac{27}{2(a-1)(a+1)^3}$ $\to A=\dfrac12a+(\dfrac{27}{2(a-1)(a+1)^3}+\dfrac12(a-1)+\dfrac12)$ $\to A\ge \dfrac12a+3\sqrt[3]{(\dfrac{27}{2(a-1)(a+1)^3}\cdot \dfrac12(a-1)\cdot \dfrac12}$ $\to A\ge \dfrac12a+\dfrac9{2(a+1)}$ $\to A\ge \dfrac12(a+1)+\dfrac9{2(a+1)}-\dfrac12$ $\to A\ge 2\sqrt{\dfrac12(a+1)\cdot\dfrac9{2(a+1)}}-\dfrac12$ $\to A\ge 3-\dfrac12$ $\to A\ge \dfrac52$ Dấu = xảy ra khi $a=2$ Bình luận
Giải thích các bước giải:
Ta có:
$A=a+\dfrac{27}{2(a-1)(a+1)^3}$
$\to A=\dfrac12a+(\dfrac{27}{2(a-1)(a+1)^3}+\dfrac12(a-1)+\dfrac12)$
$\to A\ge \dfrac12a+3\sqrt[3]{(\dfrac{27}{2(a-1)(a+1)^3}\cdot \dfrac12(a-1)\cdot \dfrac12}$
$\to A\ge \dfrac12a+\dfrac9{2(a+1)}$
$\to A\ge \dfrac12(a+1)+\dfrac9{2(a+1)}-\dfrac12$
$\to A\ge 2\sqrt{\dfrac12(a+1)\cdot\dfrac9{2(a+1)}}-\dfrac12$
$\to A\ge 3-\dfrac12$
$\to A\ge \dfrac52$
Dấu = xảy ra khi $a=2$