<s>[INST] 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.
 [/INST] 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.
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