Prove for all $a, b, c \in \mathbb {R^+}$ ${a\over\sqrt{a+b}} + {b\over\sqrt{b+c}} + {c\over\sqrt{c+a}} \gt \sqrt{a+b+c}$ is true 
Prove for all $a, b, c \in \mathbb {R^+}$ $${a\over\sqrt{a+b}} + {b\over\sqrt{b+c}} + {c\over\sqrt{c+a}} \gt \sqrt{a+b+c}$$ is true

I have been studying about inequalities and how to prove different inequalities, and I stumbled over this question which I failed to answer. I failed to see a connection between $\sqrt{a+b+c} \text{ and} {a\over\sqrt{a+b}} + {b\over\sqrt{b+c}} + {c\over\sqrt{c+a}}$. Is there any way I can prove this question and if there is can you please help. Thank you.
 A: You need only Jensen applied to convex function $f(x)={1 \over \sqrt{x}}$:
$$LHS=a\space f(a+b)+b\space f(b+c)+c\space f(c+a)\ge(a+b+c)\space f(\frac{a(a+b)+b(b+c)+c(c+a)}{a+b+c})$$
$$LHS\ge(a+b+c)\space f(\frac{(a+b+c)^2-ab-bc-ac}{a+b+c})$$
$$LHS\ge(a+b+c)\sqrt{\frac{a+b+c}{(a+b+c)^2-ab-bc-ac}}$$
$$LHS\gt(a+b+c)\sqrt{\frac{a+b+c}{(a+b+c)^2}}$$
$$LHS\gt\sqrt{a+b+c}$$
A: You may proceed as follows using convexity of $\frac{1}{\sqrt{x}}$:


*

*$\sum_{k=1}^3 \lambda_k f(x_k) \geq f\left( \sum_{k=1}^3 \lambda_k x_k\right)$ with $\sum_{k=1}^3 \lambda_k = 1$, $\lambda_1,\lambda_2,\lambda_3 \geq 0$
$$
\begin{eqnarray*} {a\over\sqrt{a+b}} + {b\over\sqrt{b+c}} + {c\over\sqrt{c+a}}
& = & (a+b+c)\sum_{cyc}\frac{a}{(a+b+c)(a+b)} \\
& \stackrel{Jensen}{\geq} & (a+b+c)\frac{1}{\sqrt{\frac{1}{a+b+c}\sum_{cyc}a(a+b)}} \\
& \color{blue}{>} & (a+b+c)\frac{1}{\sqrt{\frac{1}{a+b+c}\sum_{cyc}a(a+b\color{blue}{+c})}} \\
& = & (a+b+c)\frac{1}{\sqrt{\frac{1}{a+b+c}(a+b+c)^2}} \\ 
& = & \color{blue}{\sqrt{a+b+c}}
\end{eqnarray*}
$$
A: 
Holder's Inequality: Let $a_{ij}, 1 \leq i \leq m , 1 \leq j \leq m $ be positive real numbers. Then the following inequality holds:
$$\prod_{i = 1}^{m}\left(\sum_{j = 1}^{n}a_{ij}\right) \geq
\left(\sum_{j = 1}^{n} \sqrt[m]{\prod_{i = 1}^{m} a_{ij}}\right)^{m}.
> $$
For example, consider real numbers $a, b, c, p, q, r, x, y, z$. Then,
$$(a^{3} + b^{3} + c^{3})(p^{3} + q^{3} + r^{3})(x^{3} + y^{3} +
> z^{3}) \geq (aqx + bqy + crz)^{3}. $$

We have 
$$\sum_{\text{cyc}} \frac{a}{\sqrt{a + b}} >  \sum_{\text{cyc}} \frac{a}{\sqrt{a + 2b}}. $$
By Holder's Inequality,
$$\left(\sum_{\text{cyc}} \frac{a}{\sqrt{a + 2b}}\right)\left(\sum_{\text{cyc}} \frac{a}{\sqrt{a + 2b}}\right)\left(\sum_{\text{cyc}} a(a + 2b)\right) \geq (a + b+ c)^{3}. $$
Thus,
$$\left(\sum_{\text{cyc}}\frac{a}{\sqrt{a + 2b}}\right)^{2} \geq a + b + c, $$
and the result follows.
A: too long; can't fit into a comment. I think @Ekesh 's answer has a little flaw: @Ekesh please see if mine is correct.

Your inequality is equivalent to
$$\sum_{\text{cyc}} \frac{a}{\sqrt{a + b}}\geq a+b+c$$
By Holder's,
$$\left(\sum_{\text{cyc}} \frac{a}{\sqrt{a + b}}\right)\left(\sum_{\text{cyc}} \frac{a}{\sqrt{a + b}}\right)\left(\sum_{\text{cyc}} a(a + b)\right) \geq (a + b+ c)^{3}.$$
We notice that (by expanding),
$$\left(\sum_{\text{cyc}} a(a + b)\right) \leq (a + b+ c)^{2}.$$
It is valid to divide the above equation with this as $\fbox{1}$ It is positive $\fbox{2}$ This move removes more value at the RHS, which will not affect the sign.
Hence we get
$$\left(\sum_{\text{cyc}}\frac{a}{\sqrt{a + b}}\right)^{2} \geq a + b + c,$$
which is what we want.
