Question

CMR:

\(\frac{a^2}{a^4+1}\le\frac{1}{2}\)

Answer 1

\(\text{BĐT} \Leftrightarrow \frac{(a-1)^2(a+1)^2}{(a^2-i)(a^2+i)}\geq 0\)

Answer 2

\(\frac{a^2}{a^4+1}\le\frac{1}{2}\Leftrightarrow\left\{\begin{matrix}a^2=b\\\frac{b}{b^2+1}-\frac{1}{2}\le0\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}b\ge0\\b^2-2b+1\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{\begin{matrix}b\ge0\\\left(b-1\right)^2\ge0\end{matrix}\right.\Rightarrow\forall b\) đáp số đúng với mọi a đẳng thức chỉ xây ra khi b=1=> a={+-1}