The first attempt of a MnZn

Prove: If \(a > 0\)，then \(\forall x\in \mathbb R, a(\mathrm e^x+a)-x> 2\ln a+\dfrac32\)

Proof:

Let \(f(x) = a(\mathrm e^x+a)-x\), then \(f'(x) = a\mathrm e^x - 1\).

Since \(f'(x) > 0\) iff \(x < -\ln a\), \(\max f(x) = 1 + a ^ 2 + \ln a\).

Therefore

\[\begin{split} &\forall x\in\mathbb R\quad f(x)>2\ln a + \dfrac{3}{2} \\ \Leftrightarrow& 1+a^2+\ln a>2\ln a + \dfrac{3}{2} \\ \Leftrightarrow & a^2 > \ln a + \frac 1 2 \\ \Leftarrow &a^2 > a - \dfrac{1}{2} \end{split} \]

\(\Box\)