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

How do you prove that for all positive $a,b,c$ this formula holds true:

\begin{equation*} \frac{a}{\sqrt{a+b}} + \frac{b}{\sqrt{b+c}} + \frac{c}{\sqrt{c+a}} > \sqrt{a+b+c}? \end{equation*}

Any help will be invaluable.

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If this is homework, please consider adding the homework tag. –  Aryabhata Oct 16 '10 at 14:57

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

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

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That is what I was hinting at! Even thought you might not agree, I suggest you read this. meta.math.stackexchange.com/questions/914/…. Note these are just guidelines and in no way am I trying to force you to use them. I just want to make sure you are aware of them. –  Aryabhata Oct 16 '10 at 15:06
Moron, thanks! In this case I started my answer before your hint appeared. I thought of the last inequality first and afterwards I realized I would have to use the equality. –  Américo Tavares Oct 16 '10 at 15:24