<s>[INST] Prove that $(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$ 
Let $a,b,c,d>0$. Prove that $$(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$$

I don't know how to begin to solve this problem
 [/INST] Let $a=\min\{a,b,c,d\}$, $b=a+u$, $c=a+v$ and $d=a+w$.
Hence, $(a+b)(b+c)(c+d)(d+a)-\prod\limits_{cyc}(a+b+c-d)=$
$$4(u^2+v^2+w^2-uv-vw)a^2+(4(u^3+v^3+w^3)-2u^2v-2v^2u-2v^2w-2w^2v)a+$$
$$+\sum\limits_{cyc}(u^4-2u^2v^2+u^2vw)+u^2w^2$$
Since $\sum\limits_{cyc}(2u^3-u^2v-u^2w)=\sum\limits_{cyc}(u+v)(u-v)^2\geq0$,
it remains to prove that $\sum\limits_{cyc}(u^4-2u^2v^2+u^2vw)\geq0$, which is Schur:
$$\sum\limits_{cyc}(u^4-2u^2v^2+u^2vw)=\sum\limits_{cyc}(u^4-u^3v-u^3w+u^2vw)+\sum\limits_{cyc}uv(u-v)^2\geq0$$
Done!
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