Prove $\frac{ab}{4-d}+ \frac{bc}{4-a}+\frac{cd}{4-b}+\frac{da}{4-c} \leqslant \frac43$ for $a^2 + b^2 + c^2 + d^2 = 4$ 
Let $a,b,c,d  \geqslant 0$ and $a^2+b^2+c^2+d^2=4$. Prove that
$$\frac{ab}{4-d}+ \frac{bc}{4-a}+\frac{cd}{4-b}+\frac{da}{4-c} \leqslant \frac43.$$

I try some reverse AM-GM techniques but fail. I don't think rearrangement inequality works because of cyclic nature of this inequality. I try to homogenize this inequality to remove the constrain, but the square root in the denominator kills me.
 A: Note: in this solution, $\sum$ denotes the cyclic sum, so that $\sum f(a, b, c, d)=f(a, b, c, d)+f(b, c, d, a)+f(c, d, a, b)+f(d, a, b, c)$. 
Lemma: $\sum c^4+2\sum abc^2+6\sum ab\le 36$. 
Proof: Expanding the inequality $\sum[2(a-b)^2+(ab-1)^2]\ge 0$ gives:
$$6\sum ab\le 20+\sum a^2b^2$$. 
Expanding $\sum (ab-ac)^2\ge 0$ gives:
$$2\sum abc^2\le \sum a^2b^2+2(a^2c^2+b^2d^2)$$
Summing these, we have:
\begin{align*}
\sum c^4+2\sum abc^2+6\sum ab &\le 20+\sum c^4+2\sum a^2b^2+2(a^2c^2+b^2d^2)
\\ &= 20+(a^2+b^2+c^2+d^2)^2
\\ &= 36
\end{align*}
so the lemma is true as required. 
Now we may use the lemma, along with Cauchy Schwarz of the form $\sum \frac{x_i^2}{y_i}\ge \frac{(\sum x_i)^2}{\sum y_i}$:
\begin{align*}
\sum\frac{ab}{4-d}&= \sum\frac{2ab}{7-d^2+(d-1)^2}
\\ &\le \sum\frac{2ab}{7-d^2}
\\ &= \sum\frac{2ab}{3+2ab+c^2+(a-b)^2}
\\ &\le \sum\frac{2ab}{3+2ab+c^2}
\\ &=4-\sum \frac{c^2+3}{c^2+2ab+3}
\\ &=4-\sum \frac{(c^2+3)^2}{(c^4+2abc^2+6ab)+6c^2+9}
\\ &\le 4-\frac{(\sum c^2+12)^2}{\sum(c^4+2abc^2+6ab)+6\sum c^2+36}
\\ &\le 4-\frac{16^2}{36+60}
\\ &=4-\frac{8}{3}
\\ &=\frac{4}{3}
\end{align*}
so the inequality is proved. Equality holds at $a=b=c=d=1$.
A: Alternative proof:
We have
$$\frac{5}{18} + \frac{1}{18}d^2
- \frac{1}{4 - d} = \frac{(2 - d)(d - 1)^2}{18(4 - d)} \ge 0.$$
Thus, we have
\begin{align*}
 &\frac{ab}{4-d}+ \frac{bc}{4-a}+\frac{cd}{4-b}+\frac{da}{4-c}\\
 \le\,& \frac{5}{18}\sum_{\mathrm{cyc}} ab + \frac{1}{18}\sum_{\mathrm{cyc}}abd^2 \\
 \le\,& \frac{5}{18}\sum_{\mathrm{cyc}} \frac{a^2 + b^2}{2} + \frac{1}{18}\sum_{\mathrm{cyc}}\frac{a^2 + b^2}{2}\cdot d^2\\
 =\,& \frac{5}{18} (a^2 + b^2 + c^2 + d^2) + \frac{1}{18}
 \left(\frac{(a^2+c^2)(b^2+d^2)}{2} + a^2c^2 + b^2d^2\right)\\
 \le\,& \frac{5}{18}(a^2 + b^2 + c^2 + d^2) + \frac{1}{18}
 \left(\frac{(a^2+c^2)(b^2+d^2)}{2} + \frac{(a^2+c^2)^2}{4} + \frac{(b^2+d^2)^2}{4}\right)\\
 =\,& \frac{5}{18}(a^2 + b^2 + c^2 + d^2) + \frac{1}{18}
 \frac{(a^2 + c^2 + b^2 + d^2)^2}{4}\\
 =\,& \frac{4}{3}.
\end{align*}
We are done.
