Finding an upper bound for an expression involving exponential

Question

Given $T_{2} < T_{1}$, how to prove that for $t>0$ $$\sqrt{\left(1-e^{\frac{-t}{T_1}}\right)^{2} + e^{\frac{-2t}{T_2}}} \leq 1$$

Any hints to get started would be appreciated.

Answer 1

You can write what you want using this equivalences $$ \begin{align} & \sqrt{\left(1-e^{-\frac{t}{T_{1}}}\right)^2+e^{-\frac{2t}{T_{2}}}} \leq 1 \Longleftrightarrow \left(1-e^{-\frac{t}{T_{1}}}\right)^2+e^{-\frac{2t}{T_{2}}} \leq1 \Longleftrightarrow 1 - 2e^{-\frac{t}{T_{1}}}+ e^{-\frac{2t}{T_{1}}} + e^{-\frac{2t}{T_{2}}} \leq 1 \\ &\Longleftrightarrow 2e^{-\frac{t}{T_{1}}} - e^{-\frac{2t}{T_{1}}} - e^{-\frac{2t}{T_{2}}} \geq 0.\\ \end{align} $$ So you can set $f(t) = 2e^{-\frac{t}{T_{1}}} - e^{-\frac{2t}{T_{1}}} - e^{-\frac{2t}{T_{2}}}$, since $f(0)=0$, you only need to prove that $f'(t) \geq 0$ for all $t>0$ and you get what you want.