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

We can assume that $a+b+c+d=2$. Then the inequality becomes
$$
(1-d)(1-c)(1-b)(1-a)\le\left(1-\tfrac{c+d}2\right)\left(1-\tfrac{a+d}2\right)\left(1-\tfrac{a+b}2\right)\left(1-\tfrac{c+b}2\right)\tag{1}
$$
If any of $a$, $b$, $c$, or $d$ is greater than $1$, then the left side is negative and the inequality is trivial. So we can assume $0\le a,b,c,d\le1$.
Subsituting $a\mapsto1-a$, $b\mapsto1-b$, $c\mapsto1-c$, and $d\mapsto1-d$ shows that the inequality is equivalent to
$$
\begin{align}
abcd
&\le\left(\frac{c+d}2\right)\left(\frac{a+d}2\right)\left(\frac{a+b}2\right)\left(\frac{c+b}2\right)\\
&=\left(\frac{ac-bd}4\right)^2+\left(\frac1a+\frac1b+\frac1c+\frac1d\right)\frac{abcd}8\tag{2}
\end{align}
$$
Since $\frac1x$ is convex for $x\gt0$, Jensen's Inequality says that
$$
\begin{align}
\frac14\left(\frac1a+\frac1b+\frac1c+\frac1d\right)&\ge\frac1{\frac14(a+b+c+d)}\\
&=2\tag{3}
\end{align}
$$
$(3)$ shows that $(2)$ is true, which in turn shows that $(1)$ is true.
