# Prove that $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}} \gt 2$ [duplicate]

Prove that If $$a,b,c \in R^+$$ Then

$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}} \gt 2$$

My try:

I assumed:

$$x=\sqrt{\frac{a}{b+c}}$$

$$y=\sqrt{\frac{b}{a+c}}$$

$$x=\sqrt{\frac{c}{b+a}}$$

Then we have:

$$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)$$

Now by AM GM Inequality we have:

$$\left(\frac{a}{b}+\frac{b}{a}\right) \ge 2$$

So we get:

$$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \ge 6 \tag{1}$$

Using $$(1)$$ we need to prove:

$$x+y+z \gt 2$$

Any idea here?

• (1) does not give you $x+y+z >2$. – Kavi Rama Murthy Jan 31 '19 at 9:12
• but how can $c=0$, $a,b,c$ are positive reals – Umesh shankar Jan 31 '19 at 9:17
• @Umeshshankar Right, you should probably mention that $a, b, c > 0$, because $\mathbb{R}^+$ is a bit ambiguous whether it includes $0$. – orlp Jan 31 '19 at 9:18

Writing cyclic sums for short, $$\sum_{cyc}\sqrt{\frac{a}{b+c}}=\sum_{cyc}\frac{2a}{2\sqrt{a(b+c)}}\\ \geq\sum_{cyc}\frac{2a}{a+b+c}=2$$ where the AM-GM inequality was used for each denominator, i.e. $$2\sqrt{a(b+c)} \le a + (b+c)$$
• So is the inequality $\ge 2$? – Umesh shankar Jan 31 '19 at 9:30
• Yes. To be more precise, equality would only hold in the AM-GM terms if $a = b+c$, $b = a+c$, $c = b+a$ which would entail $a=b=c=0$ which is not possible. So we can exclude this case, and $>$ holds. – Andreas Jan 31 '19 at 9:33
• Well, what do you mean by "exceptions"? I was explaning that equality in AM-GM - as used here - holds if and only if the two terms are equal (in all three cases where AM-GM was used). Then I argued that if $a,b,c \in R^+$, this equality condition cannot occur. So I concluded that indeed $>$ holds. – Andreas Jan 31 '19 at 9:45