Find the minimum of $\sum\limits_{cyc} {\sqrt{\frac a{2(b+c)}}}$ with $a,b,c \gt 0$

Question

As said in the title, I have to find the minimum of the following:

$$\sum_{cyc} {\sqrt{\frac a{2(b+c)}}} $$ with $a+b+c>0$

In my very last attempt, I tried to work it out using AM-GM: Since $$\sqrt{\frac a{2(b+c)}}= \frac 1{\sqrt 2}\frac {a}{\sqrt{a(b+c)}}\ge \frac {\sqrt{2}a}{a+b+c}$$ $$\Rightarrow\sum_{cyc} {\sqrt{\frac a{2(b+c)}}}\ge \frac {\sqrt{2}(a+b+c)}{a+b+c}=\sqrt{2}$$ The problem is, to get the minimum value, $$a=b+c$$ $$b=c+a$$ $$c+a+b$$ or $a=b=c=0$. This is wrong since $a,b,c>0$ And after some calculus I found the minimum value is $\frac 32$ when $a=b=c$. So what mistake I have made?

Answer 1

By AM-GM $\sum\limits_{cyc}\sqrt{\frac{a}{2(b+c)}}=\sum\limits_{cyc}\frac{\sqrt2a}{2\sqrt{a(b+c)}}\geq\sum\limits_{cyc}\frac{\sqrt2a}{a+b+c}=\sqrt2$, which is infimum because we can use $c\rightarrow0^+$ and $b=c$.

Answer 2

Stronger inequality: With $a, b, c> 0$: $$\sqrt{\frac{a}{b+ c}}+ \sqrt{\frac{b}{c+ a}}+ \sqrt{\frac{c}{a+ b}}\geq \sqrt{2+ 2.\frac{abc}{\left ( a+ b \right )\left ( b+ c \right )\left ( c+ a \right )}}$$ Proof: Thus, we have: $$\left ( a+ b \right )\left ( b+ c \right )\left ( c+ a \right )+ abc= \left ( a+ b+ c \right )\left ( ab+ bc+ ca \right )$$ From the assumed inequality, we obtain the equivalent inequality: $$\sqrt{a\left ( a+ b \right )\left ( a+ c \right )}+ \sqrt{b\left ( b+ c \right )\left ( b+ a \right )}+ \sqrt{c\left ( c+ a \right )\left ( c+ b \right )}\geq 2\sqrt{\left ( a+ b+ c \right )\left ( ab+ bc+ ca \right )}$$ Other way, we have: $$\sqrt{a\left ( a+ b \right )\left ( a+ c \right )}+ \sqrt{b\left ( b+ c \right )\left ( b+ a \right )}+ \sqrt{c\left ( c+ a \right )\left ( c+ b \right )}= \sqrt{a^{2}\left ( a+ b+ c \right )+ abc}+ \sqrt{b^{2}\left ( a+ b+ c \right )+ abc}+ \sqrt{c^{2}\left ( a+ b+ c \right )+ abc}\geq \sqrt{\left ( a\sqrt{a+ b+ c}+ b\sqrt{a+ b+ c}+ c\sqrt{a+ b+ c} \right )^{2}+ 9abc}= \sqrt{\left ( a+ b+ c \right )^{3}+ 9abc}\geq 4\left ( a+ b+ c \right )\left ( ab+ bc+ ca \right )$$ Done!