Given $a,b,c,d>0$ satisfying $a+b+c+d=4$. Prove that $$\dfrac{1}{8+a^2}+\dfrac{1}{8+b^2}+\dfrac{1}{8+c^2}+\dfrac{1}{8+d^2}\leq \dfrac{4}{9}.$$
I've tried solving by assuming that $0<a\leq b\leq c \leq d$, which leads to $\dfrac{1}{8+a^2}\geq \dfrac{1}{8+b^2}\geq \dfrac{1}{8+c^2}\geq \dfrac{1}{8+d^2}$.
By using Chebyshev's sum inequality, we have $4\left(\dfrac{a}{8+a^2}+\dfrac{b}{8+b^2}+\dfrac{c}{8+c^2}+\dfrac{d}{8+d^2}\right)\leq (a+b+c+d)\left(\dfrac{1}{8+a^2}+\dfrac{1}{8+b^2}+\dfrac{1}{8+c^2}+\dfrac{1}{8+d^2}\right).$
I get nothing from this statement and really don't know that to do next.
We need to prove that $$\sum_{cyc}\left(\frac{1}{9}-\frac{1}{a^2+8}\right)\geq0$$ or $$\sum_{cyc}\frac{a^2-1}{a^2+8}\geq0$$ or $$\sum_{cyc}\left(\frac{a^2-1}{a^2+8}-\frac{2}{9}(a-1)\right)\geq0$$ or $$\sum_{cyc}\frac{(a-1)^2(7-2a)}{a^2+8}\geq0.$$ Thus, for $\max\{a,b,c,d\}\leq3.5$ our inequality is true.
Let $a\geq3.5.$
Thus, $$\sum_{cyc}\frac{1}{a^2+8}\leq\frac{1}{3.5^2+8}+\frac{3}{8}<\frac{4}{9}$$ and we are done!