Question

Chứng minh với 3 số a,b,c ta đầu có

(a-1)(a-3)(a-4)(a-6) + 9 ≥ 0

Answer 1

Ta có:

\(\left(a-1\right)\left(a-3\right)\left(a-4\right)\left(a-6\right)+9\ge0\)

\(\Leftrightarrow\left(a-1\right)\left(a-6\right)\left(a-3\right)\left(a-4\right)+9\ge0\)

\(\Leftrightarrow\left(a^2-7a+6\right)\left(a^2-7a+12\right)+9\ge0\)

Đặt \(t=a^2-7a+6\), ta có:

\(t\left(t+6\right)+9\ge0\)

\(\Leftrightarrow t^2+6t+9\ge0\)

\(\Leftrightarrow\left(t+3\right)^2\ge0\) ( hiển nhiên )

Dấu \("="\) xảy ra khi \(t=-3\)

\(\Rightarrow a^2-7a+6=-3\)

\(\Rightarrow a^2-6a-a+6=-3\)

\(\Rightarrow a\left(a-6\right)-\left(a-6\right)=-3\)

\(\Rightarrow\left(a-6\right)\left(a-1\right)=-3\)

\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{7-\sqrt{13}}{2}\\a=\dfrac{7+\sqrt{13}}{2}\end{matrix}\right.\)