$ \sqrt a+\sqrt b +\sqrt c = 1$ Prove: $\frac{a^2+bc}{\sqrt{2a^2(b+c)}}+\frac{b^2 + ac}{\sqrt{2b^2(a + c)}}+\frac{c^2 +ab}{\sqrt{2c^2(a+b)}}\geq1$ I'm having some trouble with this problem:
a, b, c are positive real numbers where $ \sqrt a + \sqrt b + \sqrt c = 1 $ .
Prove the following inequality:
$$\frac{a^2 + bc}{\sqrt{2a^2(b+c)}} + \frac{b^2 + ac}{\sqrt{2b^2(a + c)}} + \frac{c^2 +ab}{\sqrt{2c^2(a+b)}} \geq 1$$
I've tried substituting out $\geq 1$ for $\geq \sqrt a + \sqrt b + \sqrt c $ , as well as simplifying and rationalising the denominators and putting over a common denominator but I'm struggling to make it much further than that. Any help as to how to untangle this would be greatly appreciated. Many thanks :)

$\sum_{cyc}{\frac{a^2+bc}{\sqrt{2a^2(b+c)}}}\geq\sum_{cyc}{\frac{ab+ac}{\sqrt{2a^2(b+c)}}}$……①
WLOG $a\leq b\leq c$
$LHS-RHS=\sum_{cyc}{\frac{(a-b)(a-c)}{\sqrt{2a^2(b+c)}}}$
$=\frac{(c-a)(c-b)(b-a)}{\sqrt2}(\frac{1}{a(c-b)\sqrt{b+c}}-\frac{1}{b(c-a)\sqrt{a+c}}+\frac{1}{c(b-a)\sqrt{a+b}})$……②
$b(c-a)\geq c(b-a),\sqrt{a+c}\geq\sqrt{a+b}$
$\frac{1}{b(c-a)\sqrt{a+c}}\leq\frac{1}{c(b-a)\sqrt{a+b}}$
so that $②\geq0$,thus ① is established.
$\sum_{cyc}{\frac{ab+ac}{\sqrt{2a^2(b+c)}}}=\sum_{cyc}{\sqrt{\frac{b+c}{2}}}\geq\sum_{cyc}{\frac{\sqrt{b}+\sqrt{c}}{2}}=1$
