Let \(x\ge y\ge z\) be the lengths of the tangents from the vertices to the incircle and let \(R, r\) be the circumradius, inradius. Then using Euler's formula
\[f(x, y, z)=\frac{(x+y)(y+z)(z+x)}{4xyz}=\frac{R}{r}=1+\sqrt 2.\]Suppose that \(x+y\ge 2(y+z)\). Then we will show that \(f(x, y, z)>1+\sqrt 2\). Indeed, note that \(f\) is increasing in \(x\):
\[\frac{\partial }{\partial x}f(x, y, z)=\frac{(y+z)}{4yz}\left(1-\frac{yz}{x^2}\right)\ge 0.\]Thus it suffices to show that \(f(y+2z, y,z)>1+\sqrt 2\). Indeed, denoting \(t=y/z\) we need to prove that
\[\frac{(t+1)^2(t+3)}{2(t+2)t}>1+\sqrt 2\quad\iff\quad t^3+(3-2\sqrt 2)t^2+(3-4\sqrt 2)t+3>0\]for \(t\ge 1\). But it is true since for \(t=1\) it is true and also the derivative of this cubic is positive for \(t\ge 1\):
\[3t^2+(6-4\sqrt 2)t+(3-4\sqrt 2) = 3\left(t+1-\frac{2\sqrt 2}{3} \right)^2-\frac 8 3>0.\]