Elementary way of proving $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2$

Question

For all positive $a, b, c$, it holds that $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2$. (If we allow nonnegative $a, b, c$ such that at most one of them is $0$, then equality is achieved e.g. when $a=0, b=c=1$.)

This can be proven by assuming $a+b+c=1$ and defining $f(x)=\sqrt{\frac{x}{1-x}}$. Then the sum becomes $f(a)+f(b)+f(c)$. As $f$ has a single inflection point (at $x=\frac{1}{4}$,) we can apply n-1 EV to deduce that if $f(a)+f(b)+f(c)$ achieves a minimal value, then two of $a, b, c$ must be equal to each other. The rest is just calculus.

This method seems convoluted and not particularly elegant. I'm wondering if there exists a proof that doesn't apply any analytic methods (only AM-GM, Cauchy-Schwarz, etc.) If such a proof doesn't exist, what differentiates this inequality from the pretty similar Nesbitt's inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}$ that can be proved, say, with AM-HM? Which class of problems cannot be solved with elementary (olympiad) inequalities and why?

Answer 1

We have, $a,b,c>0$ or $a,b,c\in\mathbb{R}^+$.

Let $p=a+b+c>0$. Then, $p>a,b,c>0$. We can write:

$$(p-2a)^2\geq0\implies p^2\geq4a(p-a)\implies\frac{a}{p-a}\geq\frac{4a^2}{p^2}\implies\sqrt{\frac{a}{p-a}}=\sqrt{\frac{a}{b+c}}\geq\frac{2a}p\tag{1}$$

Or we can directly use the AM–GM inequality: $$\sqrt{\frac{a}{b+c}}=\sqrt{\frac{a}{p-a}}=\sqrt{\frac{a^2}{a(p-a)}}\geq\dfrac{a}{\frac{a+p-a}2}=\frac{2a}p\tag{2}$$

Similarly,

$$\sqrt{\frac{b}{p-b}}=\sqrt{\frac{b}{c+a}}\geq\frac{2b}p\tag{3}$$ $$\sqrt{\frac{c}{p-c}}=\sqrt{\frac{c}{a+b}}\geq\frac{2c}p\tag{4}$$

Summing up $(2),\ (3),$ and $(4)$, we get $$\boxed{\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\geq\frac{2(a+b+c)}p=2}\tag{5}$$

Let's verify whether equality also holds. For the equality, $p=2a=2b=2c\implies 3p=2p\implies p=0$ which contradicts $p>0$.

Hence, $$\color{red}{\boxed{\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2}}$$

Hope this helps!