We extend the algebraic Lemma used here: http://www.ams.org/amsmtgs/2259_abstracts/1143-53-201.pdf We presented this result here: https://conferinta.ssmr.ro/community
For what values of $k$ does$$(a+b+c+d)^2+kabcd\geq16+k$$hold for all $a,b,c,d>0$ satisfying $a+b+c+d=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$ ?
For $k=48$ we get the algebraic Lemma mentioned above. That Lemma was found by me in April 2018 and posted on my personal page on May 3-rd 2018. What happened next, we all know and It's not the moment to reopen old stories. But the important thing is that the story continued and the issue about the creators of these configurations has been settled.
How would you solve the problem?
By the Vasc's EV method: https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf (corollary 1.8 (a)) it's enough to understand what happens for equality case of three variables.
Now, let $c=b=a$ and $a=xd.$
Thus, the condition gives $$3a+d=\frac{3}{a}+\frac{1}{d}$$ or $$\frac{ad(3a+d)}{a+3d}=1$$ and we need to prove that $$(3a+d)^2+ka^3d\geq16+k$$ or $$\frac{x(3x+1)}{x+3}\cdot(3x+1)^2+kx^3\geq(16+k)\cdot\left(\frac{x(3x+1)}{x+3}\right)^2$$ or $$(x-1)^2((k+27)x^2-(k-18)x+3)\geq0,$$ which is obviously true for $k\leq18.$
But for $k>18$ we need $$(k-18)^2-12(k+27)\leq0,$$ which gives $k\leq48$ and we are done!