Prove that $\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$

Another way.

$$\frac{1}{ab}+\frac{1}{cd}-\frac{a^2+b^2+c^2+d^2}{2}=$$ $$=\left(\frac{1}{\sqrt{ab}}-\sqrt{ab}\right)^2+\left(\frac{1}{\sqrt{cd}}-\sqrt{cd}\right)^2+4-\left(ab+cd+\frac{a^2+b^2+c^2+d^2}{2}\right)=$$ $$=\left(\frac{1}{\sqrt{ab}}-\sqrt{ab}\right)^2+\left(\frac{1}{\sqrt{cd}}-\sqrt{cd}\right)^2+\frac{(a+b+c+d)^2}{4}-\frac{(a+b)^2+(c+d)^2}{2}=$$ $$=\left(\frac{1}{\sqrt{ab}}-\sqrt{ab}\right)^2+\left(\frac{1}{\sqrt{cd}}-\sqrt{cd}\right)^2-\frac{(a+b-c-d)^2}{4}.$$ Now, let $a+b\leq c+d$.

Thus, $0<a+b\leq2$ and by AM-GM $$\frac{1}{\sqrt{ab}}-\sqrt{ab}=\frac{1-ab}{\sqrt{ab}}\geq\frac{1-\left(\frac{a+b}{2}\right)^2}{\sqrt{ab}}\geq0.$$ Id est, it's enough to prove that: $$\frac{1}{\sqrt{ab}}-\sqrt{ab}\geq\frac{c+d-a-b}{2}$$ or $$1-ab\geq\sqrt{ab}(2-a-b)$$ or $$\sqrt{ab}(a+b)-2ab+ab-2\sqrt{ab}+1\geq0$$ or $$\sqrt{ab}(\sqrt{a}-\sqrt{b})^2+(\sqrt{ab}-1)^2\geq0$$ and we are done!