Show $a^2+b^2+c^2=1$ $\implies$ $ \dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} \le \dfrac 14 $ If $a$, $b$ and $c$ are real numbers such that $a^2+b^2+c^2=1$ , then how to prove that
$ \dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} \le \dfrac 14 $
( don't apply Schur's inequality )? Thanks.
 A: Let $x=a^2$, $y=b^2$, $z=c^2$ and $f(x,y,z)=\frac{yz}{1+x}+\frac{xz}{1+y}+\frac{xy}{1+z}$ and g(x,y,z)=x+y+z. Consider the restriction $\{(x,y,z) \in \mathbb{R}^3;g(x,y,z)=1\}$. Applying the method of Lagrange multipliers, we have
$$\frac{y}{1+z}+\frac{z}{1+y}-\frac{yz}{(1+x)^2} = \lambda$$
$$\frac{z}{1+x}+\frac{x}{1+z}-\frac{xz}{(1+y)^2} = \lambda$$
$$\frac{x}{1+y}+\frac{y}{1+x}-\frac{xy}{(1+z)^2} = \lambda$$
$$x+y+z=1$$
Subtracting the first two equations, we obtain
$$\frac{y-x}{1+z}+\frac{z}{1+y}(1-\frac{x}{1+y})-\frac{z}{1+x}(1+\frac{y}{1+x})=0 \Rightarrow$$
$$(y-x) \left( \frac{1}{1+z}+\frac{z(x+y+1)(x+y+2)}{(1+x)^2(1+y)^2} \right )=0 $$
The second factor is $\ge 0$, so $x=y$. Analogously, $y=z$. Thus, we conclude that $x=y=z=1/3$. How $f(1/3,1/3,1/3)=1/4$ and $f$ have a maximum in this restriction, we have the desired inequality.
A: By C-S
$$\sum_{cyc}\frac{b^2c^2}{1+a^2}=\sum_{cyc}\frac{b^2c^2}{a^2+b^2+a^2+c^2}\leq$$
$$\leq\sum_{cyc}\frac{b^2c^2}{4}\left(\frac{1}{a^2+b^2}+\frac{1}{a^2+c^2}\right)=\frac{1}{4}\sum_{cyc}\left(\frac{b^2c^2}{a^2+b^2}+\frac{b^2c^2}{a^2+c^2}\right)=$$
$$=\frac{1}{4}\sum_{cyc}\left(\frac{b^2c^2}{a^2+b^2}+\frac{a^2c^2}{a^2+b^2}\right)=\frac{1}{4}.$$
Done!
