<s>[INST] 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.
 [/INST] 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!
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