Show that $\frac {1}{3+x^2+y^2} + \frac {1}{3+y^2+z^2} +\frac{1} {3+x^2+z^2}\leq \frac {3}{5} . $ Let $x, y, z>0$ s.t. $x+y+z=3$.
Show that $$\frac {1}{3+x^2+y^2}  + \frac {1}{3+y^2+z^2} +\frac{1 } {3+x^2+z^2}\leq \frac {3}{5}\ . $$
My idea: $$3 + x^2 + y^2 \geq 1 + 2x+ 2y=7-2z $$
I notice that $f (t)=\frac {1}{7-2t} $  is a convex function but it's uselessness.
Also I have some troubles with the next inequality  $$\frac {a}{a^2+bc} +\frac {b}{b^2+ac}+\frac {c}{c^2+ab}\geq \frac {3}{2} \frac{a+b+c}{a^2+b^2+c^2} $$
My idea is to multiply and to apply Muirhead's inequality. But there are too much terms.
 A: The first inequality.
We need to prove that
$$\sum_{cyc}\frac{1}{3+x^2+y^2}\leq\frac{3}{5}$$ or
$$\sum_{cyc}\left(\frac{1}{3+x^2+y^2}-\frac{1}{3}\right)\leq\frac{3}{5}-1$$ or
$$\sum_{cyc}\frac{x^2+y^2}{x^2+y^2+3}\geq\frac{6}{5}.$$
Now, by C-S twice we obtain:
$$\sum_{cyc}\frac{x^2+y^2}{x^2+y^2+3}\geq\frac{\left(\sum\limits_{cyc}\sqrt{x^2+y^2}\right)^2}{\sum\limits_{cyc}(x^2+y^2+3)}=$$
$$=\frac{\sum\limits_{cyc}\left(2x^2+2\sqrt{(x^2+y^2)(x^2+z^2)}\right)}{2(x^2+y^2+z^2)+9}\geq$$
$$\geq\frac{\sum\limits_{cyc}\left(2x^2+2(x^2+yz)\right)}{2(x^2+y^2+z^2)+(x+y+z)^2}=\frac{\sum\limits_{cyc}(4x^2+2xy)}{\sum\limits_{cyc}(3x^2+xy)}\geq\frac{6}{5}$$ because the last inequality it's $$\sum_{cyc}(x-y)^2\geq0.$$
A: The second inequality.
By C-S we obtain:
$$\sum_{cyc}\frac{a}{a^2+bc}=\sum_{cyc}\frac{a^2(b+c)^2}{(a^3+abc)(b+c)^2}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}a(b+c)\right)^2}{\sum\limits_{cyc}(a^3+abc)(b+c)^2}=\frac{4(ab+ac+bc)^2}{\sum\limits_{cyc}(a^3+abc)(b+c)^2}.$$
Thus, it's enough to prove that
$$8(ab+ac+bc)^2(a^2+b^2+c^2)\geq3(a+b+c)\sum\limits_{cyc}(a^3+abc)(b+c)^2$$ or
$$\sum_{cyc}(5a^4b^2+5a^4c^2-6a^3b^3+4a^4bc-5a^3b^2c-5a^3c^2b+2a^2b^2c^2)\geq0,$$ which is obvious by Schur
$$\sum_{cyc}(a^4bc-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0$$ and Muirhead.
