Let $S$ denote the area of $\Delta$ and $a,b,c$ denote the side lengths of $\Delta$. Also, let $\alpha, \beta, \gamma$ be the angles opposite to $a,b,c$ respectively. Note that from sine law, $a = 2Rsin\alpha$, $b = 2Rsin\beta$, $c = 2Rsin\gamma$. So $$S = \frac{r}{2}(a+b+c) = \frac{r}{2} \cdot 2R(sin\alpha+sin\beta+sin\gamma) \le 3Rrsin\left(\frac{\alpha+\beta+\gamma}{3}\right) = \frac{3\sqrt{3}Rr}{2}$$by Jensen's inequality. Since $R \ge 2r$ from Euler's inequality, $R^2 + r^2 = \left(\frac{R^2}{4}+r^2\right)+\frac{3R^2}{4} \ge Rr + \frac{3Rr}{2} = \frac{5Rr}{2}$. Then, $\pi(R+r)^2 = \pi(R^2 + r^2 + 2Rr) \ge \frac{9\pi}{2}Rr > \frac{15 \sqrt{3}}{2}Rr \ge 5S$, and we are done. $\blacksquare$