let $a,b,c,d\in R$ and $ad-bc=1$,$e,f\ge \frac{1}{2}$,and $r^2=a^2+b^2+c^2+d^2,s=ac+bd$.show that $$\sqrt{e^2r^2+es}+\sqrt{f^2r^2+fs}\ge\sqrt{2}(e+f)$$
This inequality is from Geometry inequality,I try square two side,we must to prove $$e^2r^2+es+f^2r^2+fs+2\sqrt{(e^2r^2+es)(f^2r^2+fs)}\ge 2(e^2+f^2+2ef)\tag{1}$$ and use Cauchy-Schwarz inequality, we have $\sqrt{(e^2r^2+es)(f^2r^2+fs)}\ge (efr^2+\sqrt{ef}s)$ and $$r^2=(a^2+b^2)+(c^2+d^2)\ge 2\sqrt{(a^2+b^2)(c^2+d^2)}=2\sqrt{(ac+bd)^2+(ad-bc)^2}=2\sqrt{s^2+1}$$ so $$(1):LHS\ge 2(e^2+f^2+ef)\sqrt{s^2+1}+es+fs+\sqrt{ef}s$$ so we must to prove $$2(e^2+f^2+ef)\sqrt{s^2+1}+es+fs+\sqrt{ef}s\ge 2(e+f)^2$$
