Cauchy-Schwarz inequality problem with four variables

Question

Let a,b,c,d, are positive numbers. How can I prove

$a/(b+c) + b/(c+d) + c/(d+a) + d/(a+b) \ge 2$

I am unable to find a transformation which will lead me to the result. Any help would be very much appreciated.

I was proceeding as follows:

$E = a/(b+c) + b/(c+d) + c/(d+a) + d/(a+b) = a^2/(ab+ac) + b^2/(bc+bd) + c^2/(cd+ca) + d^2/(ad+bd)$

then I was using CS as:

$ E > (a+b+c+d)^2/( ab+ + bc + cd + ad + 2ac + 2bd)$

After this step I am getting lost because of the non-standard form which I have at the denominator.

I think I got it, now we can write:

$ E > (a+b+c+d)^2/( (a+ c)(b+d) + 2ac + 2bd) =E2$

since $(a+ c)^2 / 2 > 2ac $ and $(b+ d)^2 / 2 > 2bd $

We can write : $E2 > (a+b+c+d)^2/( (a+ c)(b+d) + (a+ c)^2 / 2 + (b+ d)^2 / 2) =E3$

replacing $x=a+c$ and $y=b+d$ we get

$ E3 = 2 (x+y)^2/( 2xy + x^2+ y^2) = 2$

Thus $E > 2$

Answer 1

Applying Titu's lemma, it suffices to show that (fill in the slight gap)

$$ (a+b+c+d)^2 \geq 2(ab+ac+bc+bd+cd+ac+da+bd ).$$

This is obviously true by expansion, since it becomes

$$ ( a -c)^2 + (b-d) ^2 \geq 0 .$$