Show that this function is decreasing

Question

I have to show that the function $f(x)=(\ln x)^2(1-\ e^{-\frac{t}{x}}), t>3$ is decreasing on $[\max(e^4,2t),\infty[$ and deduce that $f(x)\leq \max(16,(\log2t)^2)$, $x\geq 1$. The exercise suggests using that $e^{x}-1-2x \leq 0$ for $0 \leq x \leq 0.5$. I don't really know how to use this hint so i just tried to calculate $f'(x)$ but didn't manage to show this. Could someone help me out? Thanks a lot

Answer 1

Take derivatives with respect to $x$ to get $$\log^2x\left(-\frac{t}{x^2}e^{-t/x}\right)+\frac{2\log x}{x}(1-e^{-t/x}).$$

Answer 2

Take derivatives with respect to $x$ to get $$\log^2x\left(-\frac{t}{x^2}e^{-t/x}\right)+\frac{2\log x}{x}(1-e^{-t/x}).$$

This gives $$\frac{2x\log x-(te^{-t/x}\log^2x+2x\log xe^{-t/x})}{x^2}.$$ This is not positive provided $$2x\log x-(te^{-t/x}\log^2x+2x\log xe^{-t/x}\le 0,$$ or since $\log x>0$ (we are given that $x\ge 1,$ and the truth of the inequality is seen for $x=1$) we get $$2x\le te^{-t/x}\log x+2xe^{-t/x}.$$