Prove that: $\frac{bc}{a^2+1}+\frac{ac}{b^2+1}+\frac{ab}{c^2+1}\leq \frac{3}{4}$

Question

Given three positive numbers a,b,c satisfying $$a^2+b^2+c^2=1$$ Prove that: $$\frac{bc}{a^2+1}+\frac{ac}{b^2+1}+\frac{ab}{c^2+1}\leq \frac{3}{4}$$ The things I have done so far: $$\sum \limits_{cyc}\frac{bc}{a^2+1}=\sum \limits_{cyc}\frac{bc}{2a^2+b^2+c^2}\leq \sum \limits_{cyc}\frac{bc}{2ab+2ac}$$ $$=\sum \limits_{cyc}\frac{bc}{2a(b+c)}\leq \frac{1}{4}\sum \limits_{cyc}\frac{(b+c)^2}{2a(b+c)}=\frac{1}{4}\sum \limits_{cyc}\frac{b+c}{2a}$$ $$=\frac{1}{8}.\frac{\sum \limits_{cyc}bc(b+c)}{abc}=\frac{1}{8}.\frac{\sum \limits_{cyc}bc(b+c)+3abc}{abc}-\frac{3}{8}$$ $$=\frac{1}{8}.\frac{(a+b+c)(ab+bc+ca)}{abc}-\frac{3}{8}$$ $$\leq \frac{1}{8}.\frac{\sqrt{3(a^2+b^2+c^2)}(a^2+b^2+c^2)}{abc}-\frac{3}{8}$$ $$=\frac{\sqrt{3}}{8abc}-\frac{3}{8}$$ I don't know what to do anymore.

Answer 1

By AM-GM and C-S we obtain: $$\sum_{cyc}\frac{bc}{a^2+1}=\sum_{cyc}\frac{bc}{2a^2+b^2+c^2}\leq\frac{1}{4}\sum_{cyc}\frac{(b+c)^2}{a^2+b^2+a^2+c^2}\leq$$ $$\leq\frac{1}{4}\sum_{cyc}\left(\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\right)=\frac{1}{4}\sum_{cyc}\left(\frac{b^2}{a^2+b^2}+\frac{a^2}{b^2+a^2}\right)=\frac{3}{4}.$$