Prove that $\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$ Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that
$$\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$$
I write 
$$a^2+b^2+c^2+d^2=16-2\left(ab+cd+\left(a+b\right)\left(c+d\right)\right)$$. Then the inequality is equivalent to
$$\frac{1}{ab}+\frac{1}{cd} +ab+cd+\left(a+b\right)\left(c+d\right) \geq 8$$
But now I don't know how to change $ab,cd$ into the forms of $a+b$ and $c+d$. Moreover, from the last inequality, we have $\frac{1}{ab}+\frac{1}{cd} \geq 4$, whereas $\left(a+b\right)\left(c+d\right) \leq 4$. I cannot handle this. Please help me.
 A: 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!
A: Let $f(x)=\frac{1}{x(k-x)}-\frac{x^2+(k-x)^2}{2},$ where $0<x<k$.
Thus, $$f'(x)=-\frac{k-2x}{(kx-x^2)^2}-x-x+k=(2x-k)\left(\frac{1}{(kx-x^2)^2}-1\right)=$$
$$=\frac{(2x-k)(1-kx+x^2)(1+kx-x^2)}{(kx-x^2)^2}.$$
We see that $$1+kx-x^2=1+x(k-x)>0.$$
Consider two cases.


*

*$0<k\leq2.$
Thus, $$1-kx+x^2=\left(x-\frac{k}{2}\right)^2+1-\frac{k^2}{4}\geq0,$$ which says $$x_{min}=\frac{k}{2}$$ and
$$f(x)\geq f\left(\frac{k}{2}\right)=\frac{4}{k^2}-\frac{k^2}{4}.$$
2. $2<k<4.$
In this case we obtain $$\frac{k-\sqrt{k^2-4}}{2}<\frac{k}{2}<\frac{k+\sqrt{k^2-4}}{2},$$
which gives that $f$ gets a minimal value for $kx-x^2=1.$
Id est, $$f(x)\geq\frac{1}{1}-\frac{k^2-2}{2}=2-\frac{k^2}{2}.$$
Now, let $a+b=k\leq2.$
Thus, $c+d=4-k\geq2$ and
$$\frac{1}{ab}+\frac{1}{cd}-\frac{a^2+b^2+c^2+d^2}{2}=\frac{1}{ab}-\frac{a^2+b^2}{2}+\frac{1}{cd}-\frac{c^2+d^2}{2}\geq$$
$$\geq\frac{4}{k^2}-\frac{k^2}{4}+2-\frac{(4-k)^2}{2}=\frac{(2-k)^3(3k+2)}{4k^2}\geq0$$ and we are done!
A: $ \frac{2}{a^2+b^2} \leq \frac{1}{ab} , \frac{2}{c^2+d^2} \leq \frac{1}{cd}$
$  \Rightarrow \frac{2(a^2+b^2+c^2+d^2)}{(a^2+b^2)(c^2+d^2)} \leq \frac{1}{ab} + \frac{1}{cd}$
$  \Rightarrow  \frac{a^2+b^2+c^2+d^2}{2} \cdot (\frac{4}{(a^2+b^2)(c^2+d^2)} - 1)\leq \frac{1}{ab} + \frac{1}{cd} - \frac{a^2+b^2+c^2+d^2}{2}$
also, $ \frac{4}{(a^2+b^2)(c^2+d^2)} \geq \frac{1}{abcd} \geq 1 (\because 1 = (\frac{a+b+c+d}{4})^4 \geq abcd)$
$ \therefore  0 \leq \frac{a^2+b^2+c^2+d^2}{2} \cdot (\frac{4}{(a^2+b^2)(c^2+d^2)} - 1)\leq \frac{1}{ab} + \frac{1}{cd} - \frac{a^2+b^2+c^2+d^2}{2} , \frac{1}{ab} + \frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2} $
