91Ó°ÊÓ

Let's first define a function: $$ f(t) = e^t - (1+t)^{1+t} $$ Now, to make the inequality in terms of this function, let's take the natural logarithm for both sides of the given inequality: $$ \ln(e^t) < \ln((1+t)^{1+t}) $$ This simplifies to $$ t < (1+t)\ln(1+t) - t $$ Now subtract \(t\) from both sides and rewrite the inequality in terms of the function \(f(t)\): $$ f(t) = e^t - (1+t)^{1+t}<0 $$

Now let's analyze the properties of the function \(f(t)\) by calculating its derivative: $$ f'(t) = \frac{d}{dt}\left(e^t - (1+t)^{1+t}\right) $$ First, calculate the derivative of the exponential function: $$ \frac{d}{dt}(e^t) = e^t $$ Then, calculate the derivative of the power function using the chain rule: $$ \frac{d}{dt}((1+t)^{1+t}) = (1+t)\left(\frac{d}{dt}(1+t)\right)((1+t)^t) + \frac{d}{dt}(1+t)((1+t)^{1+t}) $$ So, $$ f'(t) = e^t - ((1+t)(1)(1+t)^t + (1)(1+t)^{1+t}) $$

Exponential function \(e^t\) is always positive, and \((1+t)^{1+t}\) is also always positive. So, their difference is negative or zero, but the inequality states that: $$ f(t) = e^t - (1+t)^{1+t}<0 $$ By analyzing the inequality, we find that \(f(t)\) is negative if \((1+t)^{1+t}>e^t\), which is the inequality we want to prove. Since \(e^t\) is strictly increasing for positive \(t\), and \((1+t)^{1+t}\) is also strictly increasing for positive \(t\), in other words, both functions are strictly increasing, we just need to show that \(f(t)\) is a strictly decreasing function. Now, recall that the derivative \(f'(t) = e^t - ((1+t)(1)(1+t)^t + (1)(1+t)^{1+t})\). For \(t>0\), \(e^t\) is strictly increasing. However, \((1+t)^{1+t}\) is also strictly increasing, but it's growing faster than \(e^t\) due to the term \((1+t)\) in front of it. Thus, \(f'(t)\) is negative for \(t>0\), therefore the function is strictly decreasing for \(t>0\).

Since we have determined that \(f(t)\) is strictly decreasing for \(t>0\), we know that if \(f(0) < 0\), the inequality would be satisfied. So now let's calculate \(f(0)\): $$ f(0) = e^0 - (1+0)^{1+0} = 1 - 1^1 = 0 $$ We know that \(f(t)\) can't be zero. Therefore, since \(f'(t)\) is strictly decreasing for \(t>0\), it must be that \(f(t)<0\) for \(t>0\). Thus, the inequality is satisfied: $$ e^t < (1+t)^{1+t} $$ for \(t>0\).