Proof of an inequality involving three numbers $(a,b,c)\gt 0$ If $(a,b,c)\gt 0$
the following inequality holds:
$$\dfrac{a^2}{2a^2+(b+c)^2}+\dfrac{b^2}{2b^2+(c+a)^2}+\dfrac{c^2}{2c^2+(a+b)^2}\lt\dfrac{2}{3}$$
I am stuck to find a proof of it.
Can someone help me? Thanks.
 A: WLOG, assume that $a\ge b \ge c$.
We first show that 
$$\dfrac{b^2}{2b^2+(c+a)^2}+\dfrac{c^2}{2c^2+(a+b)^2} \le \frac{(b+c)^2}{2(b+c)^2+a^2}.\qquad (1)$$
Indeed, it is equivalent to each of the following inequalities:
$$\dfrac{c^2}{2c^2+(a+b)^2} \le \frac{(b+c)^2}{2(b+c)^2+a^2} - \dfrac{b^2}{2b^2+(c+a)^2}$$
$$\dfrac{c^2}{2c^2+(a+b)^2} \le \frac{(b+c)^2(c+a)^2 - a^2b^2}{(2(b+c)^2+a^2)(2b^2+(c+a)^2)}$$
$$\dfrac{c^2}{2c^2+(a+b)^2} \le \frac{c(a+b+c)(c^2+ca+cb+2ab)}{(2(b+c)^2+a^2)(2b^2+(c+a)^2)}$$
$$\dfrac{c}{2c^2+(a+b)^2} \le \frac{(a+b+c)(c^2+ca+cb+2ab)}{(2(b+c)^2+a^2)(2b^2+(c+a)^2)}$$
$$\dfrac{c}{2c^2+(a+b)^2} \le \frac{c(a+b+c)^2 + 2ab(a+b+c)}{(2(b+c)^2+a^2)(2b^2+(c+a)^2)}$$
It suffices to prove that
$$\dfrac{c}{2c^2+(a+b)^2} \le \frac{c(a+b+c)^2}{(2(b+c)^2+a^2)(2b^2+(c+a)^2)}$$
or equivalently
$$\dfrac{1}{2c^2+(a+b)^2} \le \frac{(a+b+c)^2}{(2(b+c)^2+a^2)(2b^2+(c+a)^2)}$$
The last inequality is true because $2(b+c)^2 + a^2 \le (a+b+c)^2$ and $2b^2+(c+a)^2 \le 2c^2+(a+b)^2$.
Back to the original inequality. Applying $(1)$ it suffices to prove
$$\dfrac{a^2}{2a^2+(b+c)^2}+ \frac{(b+c)^2}{2(b+c)^2+a^2} \lt\dfrac{2}{3}.$$
Denote $x=a^2,y=(b+c)^2$ it is easy to show that the above inequality is equivalent to $(x-y)^2 \ge 0$.
A: A full expanding gives $\sum\limits_{sym}(\frac{1}{2}a^6+2a^5b-a^4b^2-2a^3b^3+4a^4bc+14a^3b^2c+\frac{1}{2}a^2b^2c^2)\geq0$,
which after using Schur gives $\sum\limits_{sym}(2a^5b-2a^3b^3+4a^4bc+14a^3b^2c)\geq0$, which is obvious.
