<s>[INST] Prove $\frac{1}{bc + cd + da -1} + \frac{1}{ab + cd + da -1} + \frac{1}{ab + bc + da -1} + \frac{1}{ab + bc + cd -1} \le 2$ If $a,b,c,d > 0$ and $abcd = 1$, prove that the following inequality holds:
$$\frac{1}{bc + cd + da -1} + \frac{1}{ab + cd + da -1} +  \frac{1}{ab + bc + da -1} +  \frac{1}{ab + bc + cd -1} \le 2$$

The problem I had is because it says that I need to prove that it's less than or equal to 2. With this I can't use any better known inequality (Cauchy-Schwarz, Holder, Minkowski...), because using that I'll prove that  $\frac{1}{bc + cd + da -1} + \frac{1}{ab + cd + da -1} +  \frac{1}{ab + bc + da -1} +  \frac{1}{ab + bc + cd -1}$ is bigger than something, and even if I prove that that new inequality is true, that doesn't mean that the original inequallity holds.
Other my idea was to split the 2 in  $4 \times  \frac{1}{2}$ and i'll get:
$$\frac{1}{bc + cd + da -1} + \frac{1}{ab + cd + da -1} +  \frac{1}{ab + bc + da -1} +  \frac{1}{ab + bc + cd -1} \le \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}$$
Now I need to prove that: 
$$\frac{1}{bc + cd + da -1} \le \frac{1}{2}$$
$$\frac{1}{ab + cd + da -1} \le \frac{1}{2}$$
$$\frac{1}{ab + bc + da -1} \le \frac{1}{2}$$
$$\frac{1}{ab + bc + cd -1} \le \frac{1}{2}$$ 
or:
$$bc + cd + da -1 \ge 2$$
$$ab + cd + da -1 \ge 2$$
$$ab + bc + da -1 \ge 2$$
$$ab + bc + cd -1 \ge 2$$ 
But if all inequalites hold then $a=b=c=d=1$ and that's as far I can go.
 [/INST] By inequality of arithmetic and geometric means, $bc+da\ge 2\sqrt{abcd}=2$, so
$$\frac{1}{bc+cd+da-1}\le\frac{1}{1+cd}.\tag{1}$$
Similarly, 
$$\frac{1}{ab+bc+da-1}\le\frac{1}{1+ab}.\tag{2}$$
$(1)+(2)$ gives
$$\frac{1}{bc+cd+da-1}+\frac{1}{ab+bc+da-1}\le\frac{1}{1+cd}+\frac{1}{1+ab}=\frac{2+ab+cd}{1+ab+cd+abcd}=1.\tag{3}$$
Similarly,
$$\frac{1}{ab+cd+da-1}+\frac{1}{ab+bc+cd-1}\le 1.\tag{4}$$
The conclusion follows from $(3)+(4)$.
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