Inequality for a rational function of three variables $x,y,z$ are positive real numbers such that $$x^2+y^2+z^2=1$$ Prove that $\dfrac{x^2}{1+2yz}+\dfrac{y^2}{1+2xz}+\dfrac{z^2}{1+2xy} \geqslant \dfrac{3}{5}$.Again, I try with Engel form of Cauchy inequality... 

Approaching the problem through Engel's form of Cauchy Schwarz- 
$$\dfrac{x^4}{x^2+2x^2yz}+\dfrac{y^4}{y^2+2xy^2z}+\dfrac{z^4}{z^2+2xyz^2} \geqslant  \dfrac{(x^2+y^2+z^2)^2}{x^2+y^2+z^2+2xyz(x+y+z)}$$
Now, given condition reduces the inequality to -
$$\dfrac{x^4}{x^2+2x^2yz}+\dfrac{y^4}{y^2+2xy^2z}+\dfrac{z^4}{z^2+2xyz^2} \geqslant  \dfrac{1}{1+2xyz(x+y+z)}$$
If we can prove $\dfrac{1}{1+2xyz(x+y+z)} \geq \frac35$ we are done.
It can be rewritten as $\space 5\geq 3 + 6xyz(x+y+z) \\ \iff 1\geq 3xyz(x+y+z)\\ \iff (x^2+y^2+z^2)^2\geq 3xyz(x+y+z) \\ \iff x^4+y^4+z^4 +2x^2y^2+2y^2z^2 +2z^2x^2 \geq 3xyz(x+y+z) $
The last inequality is true as from AM-GM
$$\frac{x^4+\frac{x^2y^2}{2}+\frac{x^2z^2}{2}+y^2z^2}{3}\geq x^2yz $$
