Here is my proof: let $\cot A=x, \cot B=y, \cot C=z$, then we have $xy+yz+zx=1$. And we also have \[\cos^2 A=\frac{x^2}{x^2+1},\cos^2 B=\frac{y^2}{y^2+1},\cos^2 C=\frac{z^2}{z^2+1}\] By Cauchy-Schwarz we get \[1+\cos^2{A}+\cos^2{B}=1+\frac{x^2}{x^2+1}+\frac{y^2}{y^2+1}\geq 1+\frac{(x+y)^2}{x^2+y^2+2}\] Thus we only need to show that \[\frac{x^2+y^2+2}{x^2+y^2+2+(x+y)^2}+\frac{y^2+z^2+2}{y^2+z^2+2+(y+z)^2}+\frac{z^2+x^2+2}{z^2+x^2+2+(z+x)^2}\leq 2\] which is equivalent to \[\frac{(x+y)^2}{x^2+y^2+2+(x+y)^2}+\frac{(y+z)^2}{y^2+z^2+2+(y+z)^2}+\frac{(z+x)^2}{z^2+x^2+2+(z+x)^2}\geq 1\] By Cauchy-Schwarz again \[\sum_{cyc}{\frac{(x+y)^2}{x^2+y^2+2+(x+y)^2}}\geq \frac{4(x+y+z)^2}{4(x^2+y^2+z^2)+6+2(xy+yz+zx)}=1\] which is true since $xy+yz+zx=1$. Equal holds when $A=B=C$. So the original inequality is proved .

Well done,Tourish! Here is a very nice proof..(not by me)

Very nice inequality and very nice solutions!

a^2=(bcosC+ccosB)^2≤(b^2+c^2)[(cosB)^2+(cosC)^2], \[ \frac{1}{1+\cos^{2}A+\cos^{2}B}\le \]≤(b^2+c^2)/[a^2+b^2+c^2]

chinacai wrote: a^2=(bcosC+ccosB)^2≤(b^2+c^2)[(cosB)^2+(cosC)^2], \[ \frac{1}{1+\cos^{2}B+\cos^{2}C}\le \]≤(b^2+c^2)/[a^2+b^2+c^2]

chinacai wrote: a^2=(bcosC+ccosB)^2≤(b^2+c^2)[(cosB)^2+(cosC)^2], \[ \frac{1}{1+\cos^{2}A+\cos^{2}B}\le \]≤(b^2+c^2)/[a^2+b^2+c^2] \[a^2=(bcosC+ccosB)^2\le(b^2+c^2)(cos^2B+cos^2C)\] \[\Rightarrow \frac{1}{1+\cos^{2}A+\cos^{2}B} \le \frac{b^2+c^2}{a^2+b^2+c^2}\] \[ \frac{1}{\lambda+\cos^{2}A+\cos^{2}B} \le \frac{b^2+c^2}{\lambda (a^2+b^2+c^2)}\] In $\triangle ABC$ , for $0<\lambda \le1$ ,we have \[\frac{1}{\lambda +\cos^2 A+\cos^2 B}+\frac{1}{\lambda +\cos^2 B+\cos^2 C}+\frac{1}{\lambda +\cos^2 C+\cos^2 A}\le \frac{2}{\lambda }.\] here