<s>[INST] 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.  
 [/INST] $\displaystyle \frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}\le\frac{3}{4}$ 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 $x+y+z=1$.
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