Question 38: Let the function $f(x)=x^{2}+2 x+a \cdot \ln x(a \in R)$, when $t \geq 1$, the inequality $f(2 t-1) \geq$ $2 f(t)-3$ always holds, find the range of the real number $a$.

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The function is given as $f(x)=x^{2}+2 x+a \cdot \ln x$ where $a \in R$. For $t \geq 1$, the inequality $f(2 t-1) \geq 2 f(t)-3$ must always hold. Determine the range of the real number $a$.
Question 38, Solution: In fact,
$$
\begin{array}{l}
f(2 t-1) \geq 2 f(t)-3 \\
\Leftrightarrow(2 t-1)^{2}+2(2 t-1)+a \cdot \ln (2 t-1) \geq 2\left[t^{2}+2 t+a \cdot \ln t\right]-3 \\
\Leftrightarrow a\left[\ln t^{2}-\ln (2 t-1)\right] \leq 2\left[t^{2}-(2 t-1)\right]
\end{array}
$$

We need to find all real numbers $\mathrm{a}$ such that for any $\mathrm{t} \in[1,+\infty), \quad(*)$ always holds. When $\mathrm{t}=1$, (*) is obviously true, so we only need to find the minimum possible value of $\frac{2\left[t^{2}-(2 t-1)\right]}{\ln t^{2}-\ln (2 t-1)}$ for $t \in(1,+\infty)$.

On one hand, $\lim _{t \rightarrow 1^{+}} \frac{2\left[t^{2}-(2 t-1)\right]}{\ln t^{2}-\ln (2 t-1)}=\lim _{t \rightarrow 1^{+}} \frac{2(2 t-2)}{\frac{1}{2}^{2} 2 t \frac{2}{2 t-1}}=2$; on the other hand, for any $t \in(1,+\infty)$, according to the logarithmic mean inequality: $\frac{2\left[t^{2}-(2 t-1)\right]}{\ln ^{2}-\ln (2 t-1)} \geq 2 \sqrt{t^{2}(2 t-1)}>2$. Therefore, the range of $\mathrm{a}$ that satisfies the condition is $(-\infty, 2)$.