Prove that $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{b+a}}>2$

Question

I'm struggling to show that $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{b+a}}>2$ using $\sqrt{a}\sqrt{b+c}≤\frac{a+b+c}{2}$ (which is derived from $A_m≥G_m$ for $2$).

Since there's three terms on the RHS, I tried:

$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{b+a}}≥2\sqrt{\sqrt{\frac{a}{b+c}}}\sqrt{\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{b+a}}}$, but the LHS looks monstrous and I couldn't simplify and show that $\sqrt{\sqrt{\frac{a}{b+c}}}\sqrt{\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{b+a}}}>1$.

I then tried:

$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{b+a}}=\frac{\sqrt{a(c+a)(b+a)}+\sqrt{b(c+b)(b+a)}+\sqrt{c(c+a)(b+c)}}{\sqrt{(a+b)(b+c)(c+a)}}$

$≤\frac{2\sqrt{\sqrt{a(c+a)(b+a)}}\sqrt{\sqrt{b(c+b)(b+a)}+\sqrt{c(c+a)(b+c)}}}{\sqrt{(a+b)(b+c)(c+a)}}$, but this also doesn't simplify anywhere.

Can someone help point me in the right direction?

Answer 1

Note that this is true for positive $a,b,c$ (I think you simply forgot to mention that).

Hint: try to prove that $$ \sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}. $$

Answer 2

It is not true, for example for $a=0$, $b=c=1$.