Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac{1}{2a^2+bc}+\frac{1}{2b^2+ac}+\frac{1}{2c^2+ab}\geq \frac{1}{\sqrt{a^2-ab+b^2}}+\frac{1}{\sqrt{b^2-bc+c^2}}+\frac{1}{\sqrt{c^2-ca+a^2}}$$ I have a solution
$$ \left(2 a^{2}+b c\right)\left(2 b^{2}+c a\right)-(a+b+c)^{2}\left(a^{2}-a b+b^{2}\right)=-(a-b)^{2}\left(a^{2}+b^{2}+c^{2}+3 b c\right) \leq 0 $$ From here on ($a+b+c=1$) it should be, $$ \left(2 a^{2}+b c\right)\left(2 b^{2}+c a\right) \leq a^{2}-a b+b^{2} $$ Applying AM-GM, we have $$ \begin{aligned} \frac{1}{2 a^{2}+b c}+\frac{1}{2 b^{2}+c a} & \geq \frac{2}{\sqrt{a^{2}-a b+b^{2}}} \\ \frac{1}{2 c^{2}+a b}+\frac{1}{2 b^{2}+c a} & \geq \frac{2}{\sqrt{b^{2}-b c+c^{2}}} \\ \frac{1}{2 a^{2}+b c}+\frac{1}{2 c^{2}+a b} & \geq \frac{2}{\sqrt{c^{2}-c a+a^{2}}} \end{aligned} $$
I need another solution. Thanks all
Another way, but in one line :)
$$\sum_{cyc}\frac{1}{2a^2+bc}-\sum_{cyc}\frac{1}{\sqrt{a^2-ab+b^2}}=$$ $$=\sum_{cyc}\left(\frac{a+b+c}{2a^2+bc}-\frac{2}{b+c}\right)+\sum_{cyc}\left(\frac{2}{a+b}-\frac{1}{\sqrt{a^2-ab+b^2}}\right)=$$ $$=\sum_{cyc}\frac{b^2+c^2+ab+ac-4a^2}{(2a^2+bc)(b+c)}+\sum_{cyc}\frac{2\sqrt{a^2-ab+b^2}-a-b}{(a+b)\sqrt{a^2-ab+b^2}}=$$ $$=\sum_{cyc}\tfrac{(c-a)(2a+c)-(a-b)(2a+b)}{(2a^2+bc)(b+c)}+\sum_{cyc}\tfrac{3(a-b)^2}{(a+b)\sqrt{a^2-ab+b^2}(2\sqrt{a^2-ab+b^2}+a+b)}=$$ $$=\sum_{cyc}(a-b)\left(\tfrac{2b+a}{(2b^2+ac)(a+c)}-\tfrac{2a+b}{(2a^2+bc)(b+c)}\right)+\sum_{cyc}\tfrac{3(a-b)^2}{2(a+b)(a^2-ab+b^2)+(a+b)^2\sqrt{a^2-ab+b^2}}=$$ $$=\sum_{cyc}\tfrac{(a-b)^2(2a^2b+2ab^2+3abc-2c^2a-2c^2b)}{(a+c)(b+c)(2b^2+ac)(2c^2+ab)}+\sum_{cyc}\tfrac{3(a-b)^2}{2(a+b)(a^2-ab+b^2)+(a+b)^2\sqrt{a^2-ab+b^2}}\geq$$ $$\geq\sum_{cyc}\tfrac{(a-b)^2(abc-2c^2a-2c^2b)}{(a+c)(b+c)(2b^2+ac)(2c^2+ab)}+\sum_{cyc}\tfrac{3(a-b)^2}{2(a+b)(a^2-ab+b^2)+(a+b)^3}=$$ $$=\sum_{cyc}\tfrac{(a-b)^2(abc-2c^2a-2c^2b)}{(a+c)(b+c)(2b^2+ac)(2c^2+ab)}+\sum_{cyc}\tfrac{(a-b)^2}{(a+b)(a^2+b^2)}=$$ $$=\sum_{cyc}\tfrac{(a-b)^2((abc-2c^2a-2c^2b)+(a+c)(b+c)(2a^2+bc)(2b^2+ac))}{(a+c)(b+c)(a+b)(a^2+b^2)(2b^2+ac)(2c^2+ab)}\geq$$ $$\geq\sum_{cyc}\tfrac{(a-b)^2((abc-2c^2a-2c^2b)+c(a+b+c)(2a^2+bc)(2b^2+ac))}{(a+c)(b+c)(a+b)(a^2+b^2)(2b^2+ac)(2c^2+ab)}=$$ $$=\sum_{cyc}\tfrac{(a-b)^2c(ab(a+b)(a^2+4ab+b^2)-2abc(a^2+b^2)+c^2(a+b)(2a^2-ab+2b^2)+abc^3)}{(a+c)(b+c)(a+b)(a^2+b^2)(2b^2+ac)(2c^2+ab)}\geq$$ $$\geq\sum_{cyc}\tfrac{(a-b)^2c(ab(a+b)(a^2+b^2)-2abc(a^2+b^2)+(a+b)(a^2+b^2)c^2)}{(a+c)(b+c)(a+b)(a^2+b^2)(2b^2+ac)(2c^2+ab)}=$$ $$=\sum_{cyc}\tfrac{(a-b)^2c(a^2+b^2)(ab(a+b)-2abc+(a+b)c^2)}{(a+c)(b+c)(a+b)(a^2+b^2)(2b^2+ac)(2c^2+ab)}\geq0.$$