How to prove that $\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq\frac{3\sqrt{3}}{4}$? Let $a,b,c>0: (a+b)(b+c)(c+a)=ab+bc+ca$. How to prove that $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq\frac{3\sqrt{3}}{4}$$
 A: Let $x=\frac{a}{\sqrt{(a+b)(a+c)}}$ $y=\frac{b}{\sqrt{(b+a)(b+c)}}$ and $z=\frac{c}{\sqrt{(c+a)(c+b)}}$. 
Then we have  $x^{2}+y^{2}+z^{2}+2xyz=1$. Note that for angles $A$, $B$ and $C$ in some acute triangle:$\cos^{2}A+\cos^{2}B+\cos^{2}C+2\cos A\cos B\cos C=1$. Hence $x$,$y$ and $z$ can be replaced by $\cos A$, $\cos B$ and $\cos C$ respectively and we have following equations (Note $(a+b)(b+c)(c+a)=ab+bc+ca$):
$\cos A=x=\frac{a}{\sqrt{(a+b)(a+c)}}$, $\sin A=\sqrt{b+c}$; $\cos B=y=\frac{b}{\sqrt{(b+a)(b+c)}}$, $\sin B=\sqrt{a+c}$; $\cos C=z=\frac {c}{\sqrt{(c+a)(c+b)}}$, $\sin C=\sqrt{a+b}$.
Now $\sum_{cyc}\frac{a}{\sqrt{a+b}}\leq\frac{3\sqrt{3}}{4}\Longleftrightarrow\sum_{cyc}\sin A\cos B\leq\frac{3\sqrt{3}}{4}$. 
Since the inequality is symmetric, we assume that $a\leq b\leq c$ without loss of generality. Then $\sin A\leq \sin B \leq \sin C$ and $\cos A\geq \cos B \geq \cos C$. By rearrangement inequality and AM–GM inequality:
$\sin A\cos B+\sin B\cos C+\sin C\cos A\leq \sin A\cos C+\sin B\cos B+\sin C\cos A=\sin B+\sin B\cos B=4\sin\frac{B}{2}\cos^{3}\frac{B}{2}\leq\frac{3\sqrt{3}}{4}$
Added by Alex Ravsky 
It seems you can finish your solution as follows. 
Let X be the smallest among angles $A,$ $B,$ and $C$ and $Z$ be the largest. 
Then $\sin X\leq \sin Y \leq \sin Z$ and $\cos X\geq \cos Y \geq \cos Z$. By rearrangement inequality and AM–GM inequality:
$\sin A\cos B+\sin B\cos C+\sin C\cos A\leq \sin X\cos Z+\sin Y\cos Y+\sin Z\cos X=\sin Y+\sin Y\cos Y=4\sin\frac{Y}{2}\cos^{3}\frac{Y}{2}\leq\frac{3\sqrt{3}}{4}.$ 
A: By Cauchy-Schwarz $\left(\sum\limits_{cyc}\frac{a}{\sqrt{a+b}}\right)^2\leq\sum\limits_{cyc}\frac{a}{(a+b)(a+b+2c)}\sum\limits_{cyc}a(a+b+2c)$.
Hence, it remains to prove that $\frac{27(a+b)(a+c)(b+c)}{16(ab+ac+bc)\sum\limits_{cyc}(a^2+3ab)}\geq\sum\limits_{cyc}\frac{a}{(a+b)(a+b+2c)}$, which after full expanding gives
$\sum\limits_{cyc}(6a^7b^2+22a^7c^2+9a^6b^3+89a^6c^3+29a^5b^4+109a^5c^4+28a^7bc+59a^6b^2c+155a^6c^2b-42a^5b^3c+246a^5c^3b+46a^4b^4c+66a^5b^2c^2-359a^4b^3c^2-231a^4c^3b^2-232a^3b^3c^3)\geq0$, 
which is true because by AM-GM easy to show that 
$288\sum\limits_{cyc}a^5c^3b\geq288\sum\limits_{cyc}a^4b^3c^2$,
$16\sum\limits_{cyc}a^7c^2\geq16\sum\limits_{cyc}a^4c^3b^2$,
$80\sum\limits_{cyc}a^6c^3\geq80\sum\limits_{cyc}a^4c^3b^2$,
$96\sum\limits_{cyc}a^6c^2b\geq96\sum\limits_{cyc}a^4c^3b^2$,
$80\sum\limits_{cyc}a^5c^4\geq80\sum\limits_{cyc}a^3b^3c^3$ and 
$32\sum\limits_{cyc}a^4c^3b^2\geq32\sum\limits_{cyc}a^3b^3c^3$.
Thus, it remains to prove that
$\sum\limits_{sym}(6a^7b^2+9a^6b^3+29a^5b^4+14a^7bc+59a^6b^2c-42a^5b^3c+23a^4b^4c+33a^5b^2c^2-71a^4b^3c^2-60a^3b^3c^3)\geq0$,
which is true by Muirhead. 
