Prove $\displaystyle\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{d+3}+\frac{d}{a+3}\le 1$. Given $a,b,c,d\ge 0$ and $a^2+b^2+c^2+d^2=4$ show the following holds :
$$\displaystyle\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{d+3}+\frac{d}{a+3}\le 1$$
Now I tried to fully expand but that becomes too ugly. I want a non-expand proof of this fact. For some motivation I solved a three variable version which I assumed to be true and it came out to be true!
That one is :
$a,b,c\ge 0$ and $a^2+b^2+c^2=3$ show that :
$\displaystyle \frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2}\le 1$. The second one is easier and I DO NOT NEED a solution of the second one. Please help me out on the first one because it is similar but not solvable by a similar method.
You can treat the function 
$$
f(a,b,c,d) = \frac{a}{b+3} = \frac{b}{c+3} + \frac{c}{d+3} + \frac{d}{a+3}
$$
as a 4-d surface, and use calculus to find any critical points.  Finding the partial derivatives of $f$ with respect to $a,b,c,d$ end up looking very symmetrical, and you end up finding that (1,1,1,1) is a critical point.  Now, you can check that this satisfies your quadratic condition, and you can either calculate the second derivatives (inadvisable), or do a local test (not a complete proof, but easier) to confirm that this point is indeed a local maximum.
Since at this local maximum, your inequality is satisfied, and you can check that there are no other critical points, ( $(-1,-1,-1,-1)$ technically works, but this is outside the given constraints), you can conclude that this inequality holds.

Addendum:  techincally speaking you should check the boundaries before you can make the conclusion.  The boundary conditions actually devolve into variants of the three term case for which you have already solved.  So everything should work out peachy there as well.
You can treat the function 
$$
f(a,b,c,d) = \frac{a}{b+3} = \frac{b}{c+3} + \frac{c}{d+3} + \frac{d}{a+3}
$$
as a 4-d surface, and use calculus to find any critical points.  Finding the partial derivatives of $f$ with respect to $a,b,c,d$ end up looking very symmetrical, and you end up finding that (1,1,1,1) is a critical point.  Now, you can check that this satisfies your quadratic condition, and you can either calculate the second derivatives (inadvisable), or do a local test (not a complete proof, but easier) to confirm that this point is indeed a local maximum.
Since at this local maximum, your inequality is satisfied, and you can check that there are no other critical points, ( $(-1,-1,-1,-1)$ technically works, but this is outside the given constraints), you can conclude that this inequality holds.

Addendum:  techincally speaking you should check the boundaries before you can make the conclusion.  The boundary conditions actually devolve into variants of the three term case for which you have already solved.  So everything should work out peachy there as well.