# Inequality of cyclic expression

For $$a,b,c,d$$ positive given that $$abcd=1.$$ Look at cyclic expression (i.e. rotating values in order that of a-b-c-d-a doesn't affect the result of equation) $$E$$ s.t. $$E=\sum_{abcd}\frac{a}{da+a+1}$$ Prove that inequality $$1 holds.

What's on my hands so far : AM-GM-HM , Schwarz , Rearrangement principle , Normalizing or Substitution , (doesn't seem so but if necessary) Jensen and Hölder

I think multiple of these techniques should be applied together here but it is quite a work for me to figure out right combination and strategy.

I can rewrite the expression in two more distinct ways : $$E = \sum_{abcd}\frac{1}{bcd+d+1} = \sum_{abcd}\frac{abc}{abc+bc+1}$$

I tried to reduce number of variables but it occurs that symmetry destructs and comes after weird expression that doesn't look any better.

I tried casework, since it is assumable to have $$max(a,b,c,d)\geq 1$$ or $$min(a,b,c,d)\leq 1$$ But not much further step was made either.

I had drawn graph of this function $$E(a,b,x)$$ and intuitively concluded that E is convex , increasing But lacking multivar-calculus knowledges i can't try to find actual minimums and maximums of E in a given range.

What approach should i take?

Let $$a=\frac{x}{y},$$ $$b=\frac{y}{z}$$ and $$c=\frac{z}{t},$$ where $$x$$, $$y$$, $$z$$ and $$t$$ are positives.
Thus, the condition gives $$d=\frac{t}{x}$$ and by C-S $$\sum_{cyc}\frac{a}{da+a+1}=\sum_{cyc}\frac{\frac{x}{y}}{\frac{t}{y}+\frac{x}{y}+1}=\sum_{cyc}\frac{x}{x+y+t}=4-\sum_{cyc}\left(\frac{x}{x+y+t}-1\right)=$$ $$=4-(y+t)\left(\frac{1}{x+y+t}+\frac{1}{z+y+t}\right)-(x+z)\left(\frac{1}{y+x+z}+\frac{1}{t+x+z}\right)\leq$$ $$\leq4-\frac{4(y+t)}{x+z+2(y+t)}-\frac{4(x+z)}{y+t+2(x+z)}.$$ Now, let $$x+z=k(y+t).$$
Id est, it's enough to prove that $$4-\frac{4}{k+2}-\frac{4k}{2k+1}<2$$ or $$\frac{1}{k+2}+\frac{k}{2k+1}>\frac{1}{2}$$ or $$3k>0.$$
Also, we have $$\sum_{cyc}\frac{a}{da+a+1}=\sum_{cyc}\frac{x}{x+y+t}>\sum_{cyc}\frac{x}{x+y+z+t}=1.$$