Prove $\frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}\le\frac{3}{4}$ if $a^2+b^2+c^2=1$ 
Ff $a,b,c$ are positive real numbers that $a^2+b^2+c^2=1$ ,Prove: $$\frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}\le\frac{3}{4}$$
  Additional info: I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.

Things I have done: I tried to  change LHS to something more easy to work but I was not successful. For example $$\frac{ab}{1+c^2}=\frac{1}{2}\left(\frac{a^2+b^2+2ab}{1+c^2}-\frac{a^2+b^2}{1+c^2}\right)$$  
that was not useful. Any hint for starting step is appreciated.  
 A: Write $x=a^2$, $y=b^2$, and $z=c^2$.  Then, $x+y+z=1$. A first consequence of this is
$$
(1+z)\geq 2\sqrt{xy+z}.\tag{*}
$$
This is true because
$$
(1+z)^2-(2\sqrt{xy+z})^2=(1+z)^2-4z-4xy\\
=(1-z)^2-4xy=(x+y)^2-4xy=(x-y)^2\geq 0.
$$
Analogous to (*), we also have
$$
(1+y)\geq2\sqrt{xz+y},\quad (1+x)\geq 2\sqrt{yz+x}.
$$
This implies
$$
\sum_{\text{cyc}}\frac{ab}{1+c^2}=\sum_{\text{cyc}}\frac{\sqrt{xy}}{1+z}\leq\frac{1}{2}\sum_{\text{cyc}}\frac{\sqrt{xy}}{\sqrt{xy+z}}.
$$
So the claim follows if we can show $\sum_{\text{cyc}}\frac{\sqrt{xy}}{\sqrt{xy+z}}\leq\frac{3}{2}$. But that has already been done here, so we are done!
p.s. For completeness, I'll produce the argument from the link here: first, observe
$$
xy+z=xy+(1-x-y)=(1-x)(1-y)=(y+z)(x+z).
$$
Then, it follows from the AM-GM inequality that
$$
\sum_{\text{cyc}}\frac{\sqrt{xy}}{\sqrt{xy+z}}=\sum_{\text{cyc}}\frac{\sqrt{xy}}{\sqrt{(x+z)(y+z)}}\leq\sum_{\text{cyc}}\frac{1}{2}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)=\frac{3}{2}.
$$
A: $\displaystyle \frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}\le\frac{3}{4}$ is equivalent to
EDIT: Do not pursue this method. It is wrong! Left as warning.
$\displaystyle \frac{a^2+b^2}{1+c^2}+\frac{b^2+c^2}{1+a^2}+\frac{c^2+a^2}{1+b^2}\le\frac{3}{2}$, as $ab\leq\frac{1}{2}(a^2+b^2)$.
This is equivalent to:
$\displaystyle \frac{1-c^2}{1+c^2}+\frac{1-b^2}{1+b^2}+\frac{1-a^2}{1+a^2} \leq \frac{3}{2}$, where $a^2+b^2+c^2=1$.
Let $x=a^2,y=b^2,z=c^2$, for simplicity.
Then the above is equivalent to
$\displaystyle \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z} \leq \frac{9}{4}$, whre $x+y+z=1$.
