It is equivalent to: $\sqrt{\frac{a^{4}+c^{4}}{2}}+\sqrt{\frac{a^{4}+d^{4}}{2}}+\sqrt{\frac{b^{4}+c^{4}}{2}}+\sqrt{\frac{b^{4}+d^{4}}{2}}\ge 2(ad+bc) $ By QM-AM and AM-GM we have: $LHS \geq \frac{a^2+c^2}{2} + \frac{a^2+d^2}{2} + \frac{b^2+c^2}{2} + \frac{b^2+d^2}{2} = (a^2+d^2)+(b^2+c^2) \geq 2ad+2bc = RHS$

it is easy if we use QM-AM and AM -GM

Turkey National Olympiad Second Round 2005

nice problem

For another approach,we can use C-S here. $\sqrt{a^4+c^4}+\sqrt{a^4+d^4}+\sqrt{b^4+c^4}+\sqrt{b^4+d^4} \ge \frac{a^2+c^2+a^2+d^2+b^2+c^2+b^2+d^2}{\sqrt2}=\sqrt{2}(a^2+b^2+c^2+d^2)$ Then we can use AM-GM we get, $\sqrt{2}(a^2+b^2+c^2+d^2) \geq 2\sqrt{2}\sqrt{(a^2+b^2)(c^2+d^2)}=2\sqrt{2}\sqrt{a^2+b^2}\sqrt{c^2+d^2}$ by C-S, $2\sqrt{2}\sqrt{a^2+b^2}\sqrt{c^2+d^2} \geq 2\sqrt{2}(ad+bc)$ as desired.