Dtjytyk

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
$$a^2+b^2+c^2+d^2=16-2left(ab+cd+left(a+bright)left(c+dright)right)$$. Then the inequality is equivalent to
$$frac{1}{ab}+frac{1}{cd} +ab+cd+left(a+bright)left(c+dright) geq 8$$
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 $left(a+bright)left(c+dright) leq 4$. I cannot handle this. Please help me.

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 $a+bleq c+d$.

Thus, $0<a+bleq2$ and by AM-GM $$frac{1}{sqrt{ab}}-sqrt{ab}=frac{1-ab}{sqrt{ab}}geqfrac{1-left(frac{a+b}{2}right)^2}{sqrt{ab}}geq0.$$
Id est, it's enough to prove that:
$$frac{1}{sqrt{ab}}-sqrt{ab}geqfrac{c+d-a-b}{2}$$ or
$$1-abgeqsqrt{ab}(2-a-b)$$ or
$$sqrt{ab}(a+b)-2ab+ab-2sqrt{ab}+1geq0$$ or
$$sqrt{ab}(sqrt{a}-sqrt{b})^2+(sqrt{ab}-1)^2geq0$$ and we are done!

$ 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} $

Let $f(x)=frac{1}{x(k-x)}-frac{x^2+(k-x)^2}{2},$ where $0<x<k$.

Thus, $$f'(x)=-frac{k-2x}{(kx-x^2)^2}-x-x+k=(2x-k)left(frac{1}{(kx-x^2)^2}-1right)=$$
$$=frac{(2x-k)(1-kx+x^2)(1+kx-x^2)}{(kx-x^2)^2}.$$
We see that $$1+kx-x^2=1+x(k-x)>0.$$
Consider two cases.

1. $0<kleq2.$

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
$$f(x)geq fleft(frac{k}{2}right)=frac{4}{k^2}-frac{k^2}{4}.$$
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, $$f(x)geqfrac{1}{1}-frac{k^2-2}{2}=2-frac{k^2}{2}.$$
Now, let $a+b=kleq2.$

Thus, $c+d=4-kgeq2$ and
$$frac{1}{ab}+frac{1}{cd}-frac{a^2+b^2+c^2+d^2}{2}=frac{1}{ab}-frac{a^2+b^2}{2}+frac{1}{cd}-frac{c^2+d^2}{2}geq$$
$$geqfrac{4}{k^2}-frac{k^2}{4}+2-frac{(4-k)^2}{2}=frac{(2-k)^3(3k+2)}{4k^2}geq0$$ and we are done!