TITLE: How to prove that $\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq\frac{3\sqrt{3}}{4}$?
QUESTION [13 upvotes]: Let $a,b,c>0: (a+b)(b+c)(c+a)=ab+bc+ca$. How to prove that $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq\frac{3\sqrt{3}}{4}$$

REPLY [2 votes]: 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.