If $x^2+y^2+z^2=3/2$, then $\sum\frac{x^2}{x(4x-3)+z^2+y^2} \le1+\frac{\sqrt2}2$ If $x$, $y$, and $z$ are real numbers satisfying 
$$
x^2 +y^2 +z^2 = 3/2
$$
then prove that $1+\frac{\sqrt{2}}{2} \geq$ the cyclic sum of 
$$
\frac{x^2}{x(4x-3)+z^2+y^2} .
$$
I've tried Cauchy-Schwarz, but I don't know how to restructure the cyclic sum to relate it to $1+\frac{\sqrt{2}}{2}$. Is there a way to do it using just basic inequalities such as Cauchy-Schwarz?
 A: I think your constant should be $1+\frac{\sqrt2}{2} \lt 1+\frac{2}{\sqrt2}$ which is stronger and still true. Since the denominator $4x^2+y^2+z^2-3x$ is bigger if $x$ is negative it is enough to prove that the inequality holds for positive real numbers $x,y,z$. Therefore, to make it a bit more beautiful, you can make the substitutions $x^2=\frac{3}{2}a$, $y^2=\frac{3}{2}b$ and $z^2=\frac{3}{2}c$. Then, the inequality changes to:
$$
\sum_{cyc} \frac{a}{4a+b+c-\sqrt{6a}} \le 1+\frac{\sqrt2}{2}
$$
For positive reals $a,b,c$ with $a+b+c=1$. Now, with AM-GM we have $\sqrt{6a}=\sqrt{(\sqrt{2})\cdot(3\sqrt{2}a)}\le\frac{1}{2}(\sqrt{2}+3\sqrt{2}a) \iff -\frac{1}{2}(\sqrt{2}+3\sqrt{2}a)\le-\sqrt{6a}$
And therefore:
$$
\sum_{cyc} \frac{a}{4a+b+c-\sqrt{6a}} \le \sum_{cyc} \frac{a}{4a+b+c-\frac{1}{2}(\sqrt{2}+3\sqrt{2}a)} = \sum_{cyc} \frac{a}{4a+b+c-\frac{1}{2}(\sqrt{2}(a+b+c)+3\sqrt{2} a)}=\sum_{cyc} \frac{a}{4a+b+c-\frac{1}{2}(4\sqrt{2}a+\sqrt{2}b +\sqrt{2} c)}=\sum_{cyc} \frac{a}{(1-\frac{\sqrt{2}}{2})(4a+b+c)}
$$
So it remains to prove that:
$$
\sum_{cyc} \frac{a}{(1-\frac{\sqrt{2}}{2})(4a+b+c)} \le 1+\frac{\sqrt2}{2} \iff \sum_{cyc} \frac{a}{4a+b+c} \le \frac{1}{2}
$$
If you multiply the whole inequality with $1-\frac{\sqrt{2}}{2}$. Now you can modify it as follows:
$$
\sum_{cyc} \frac{a}{4a+b+c} = \sum_{cyc} \frac{a}{2a+(a+b)+(a+c)} \le \sum_{cyc} \frac{a}{2a+2\sqrt{ab}+2\sqrt{ac}}=\frac{1}{2}\sum_{cyc} \frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=\frac{1}{2}
$$
Which follows again from AM-GM, with equality only if $a=b=c=\frac{1}{3}$ so in the original inequality, if $x=y=z=\frac{1}{\sqrt{2}}$.
