\(a^4+1\ge a\left(a^2+1\right)\Leftrightarrow a^3\left(a-1\right)-\left(a-1\right)\ge0\) \(\Leftrightarrow\left(a^3-1\right)\left(a-1\right)\ge0\)
\(\Leftrightarrow\left(a-1\right)^2\left(a^2+a+1\right)\ge0\) (Luôn đúng)
\(a^4+1\ge a\left(a^2+1\right)\\ \Leftrightarrow a^4+1\ge a^3+a\\ \Leftrightarrow a^4+1-a-a^3\ge0\Leftrightarrow\left(a^4-a\right)-\left(a^3-1\right)\ge0\\ \Leftrightarrow a\left(a-1\right)\left(a^2+a+1\right)-\left(a-1\right)\left(a^2+a+1\right)\ge0\\ \Leftrightarrow\left(a-1\right)\left(a^3+a^2+a-a^2-a-1\right)\ge0\\ \Leftrightarrow\left(a-1\right)\left(a^3-1\right)\ge0\\ \Leftrightarrow\left(a-1\right)\left(a-1\right)\left(a^2+a+1\right)\ge0\)
Nhận xét:
(a - 1)(a - 1) = (a - 1)2 \(\ge\) 0 ∀ x
\(a^2+a+1\\ =a^2+2.a.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(a+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy \(\left(a-1\right)^2\left(a^2+a+1\right)\) luôn dương
=> ... (đpcm)
