$\frac{a}{\sqrt{a+b}} + \frac{b}{\sqrt{b+c}} + \frac{c}{\sqrt{c+a}} > \sqrt{a+b+c}$ is true for positive a,b,c

Question

How to prove that for all positive a,b,c this formula holds true:

$\frac{a}{\sqrt{a+b}} + \frac{b}{\sqrt{b+c}} + \frac{c}{\sqrt{c+a}} > \sqrt{a+b+c}$

Any help will be invaluable.

Answer 1

Since

$$a+b+c=\sqrt{(a+b+c)^2}=\sqrt{a+b+c}\ \sqrt{a+b+c}\iff \dfrac{a+b+c}{\sqrt{a+b+c}}=\sqrt{a+b+c}$$

and

$$\dfrac{a}{\sqrt{a+b}}>\dfrac{a}{\sqrt{a+b+c}}$$

$$\dfrac{b}{\sqrt{b+c}}>\dfrac{b}{\sqrt{a+b+c}}$$

$$\dfrac{c}{\sqrt{c+a}}>\dfrac{c}{\sqrt{a+b+c}}$$

we have

$$\dfrac{a}{\sqrt{a+b}}+\dfrac{b}{\sqrt{b+c}}+\dfrac{c}{\sqrt{c+a}}>\dfrac{a+b+c}{% \sqrt{a+b+c}}=\sqrt{a+b+c}.$$