Hello. We have $a^2+3=a^2+1+1+1\geq 4\sqrt{a}\Rightarrow \sqrt{\frac{b}{a^2+3}}\leq \sqrt{\frac{b}{4\sqrt{a}}}=\frac{1}{2}\sqrt[4]{\frac{b^2}{a}}$. Thus,it suffices to show that $\sum_{cyc} \sqrt[4]{\frac{b^2}{a}}\leq \frac{3}{\sqrt[4]{abc}}\Leftrightarrow \left(\sum_{cyc} \sqrt[4]{\frac{b^2}{a}}\right)^4\leq \frac{81}{abc}$. However,Holder gives $3\left(\sum_{cyc} \frac{1}{a}\right)\cdot (b+c+a)^2\geq \left(\sum_{cyc} \sqrt[4]{\frac{b^2}{a}}\right)^4\Leftrightarrow \left(\sum_{cyc} \sqrt{\frac{b^2}{a}}\right)^4\leq 27\sum_{cyc} \frac{1}{a}$. Hence,it suffices to show that $\sum_{cyc} \frac{1}{a}\leq \frac{3}{abc}\Leftrightarrow \frac{ab+bc+ca}{abc}\leq \frac{3}{abc}\Leftrightarrow ab+bc+ca\leq 3$, which is true,since $3(ab+bc+ca)\leq (a+b+c)^2=9$.

cjquines0 wrote: Let $a,b,c$ be positive real numbers with $a+b+c=3$. Prove that \[ \sqrt{\frac{b}{a^2+3}}+ \sqrt{\frac{c}{b^2+3}}+ \sqrt{\frac{a}{c^2+3}} ~\le~ \frac32\sqrt[4]{\frac{1}{abc}}\] By AM-GM, C-S and Vasc $\sum_{cyc}\sqrt{\frac{b}{a^2+3}}\leq\frac{1}{2}\sum_{cyc}\sqrt[4]{\frac{b^2}{a}}\leq\frac{1}{2}\sqrt{3\sum_{cyc}\sqrt{\frac{b^2}{a}}}=\frac{1}{2\sqrt[4]{abc}}\sqrt{3\sum_{cyc}\sqrt{b^3c}}\leq\frac{3}{2\sqrt[4]{abc}}$.

The following inequality holds true: Let a,b,c be positive numbers such that a+b+c=3. Prove that: $\sqrt {\frac{{bc}}{{{a^2} + 3}}} + \sqrt {\frac{{ca}}{{{b^2} + 3}}} + \sqrt {\frac{{ab}}{{{c^2} + 3}}} \le \frac{3}{2}\sqrt[4]{{\frac{1}{{abc}}}}$.

cjquines0 wrote: Let $a,b,c$ be positive real numbers with $a+b+c=3$. Prove that \[ \sqrt{\frac{b}{a^2+3}}+ \sqrt{\frac{c}{b^2+3}}+ \sqrt{\frac{a}{c^2+3}} ~\le~ \frac32\sqrt[4]{\frac{1}{abc}}\] My solution Let $S = \sqrt {\frac{b}{{{a^2} + 3}}} + \sqrt {\frac{c}{{{b^2} + 3}}} + \sqrt {\frac{a}{{{c^2} + 3}}} $. Since a+b+c=3 we have $9 = {\left( {a + b + c} \right)^2} \ge 3\left( {ab + bc + ca} \right) \Rightarrow ab + bc + ca \le 3 = a + b + c$. We have ${a^2} + 3 = {a^2} + 1 + 1 + 1 \ge 4\sqrt a $ hence $\begin{array}{l} S \le \frac{1}{2}\sum {\sqrt {\frac{b}{{\sqrt a }}} } \Rightarrow {S^2} \le \frac{1}{4}{\left( {\sum {\sqrt {\frac{b}{{\sqrt a }}} } } \right)^2} \le \frac{1}{4}\left( {a + b + c} \right)\left( {\frac{1}{{\sqrt a }} + \frac{1}{{\sqrt b }}\frac{1}{{\sqrt c }}} \right) = \frac{3}{4}\left( {\frac{1}{{\sqrt a }} + \frac{1}{{\sqrt b }}\frac{1}{{\sqrt c }}} \right)\\ \Rightarrow {S^4} \le \frac{9}{{16}}{\left( {\frac{1}{{\sqrt a }} + \frac{1}{{\sqrt b }}\frac{1}{{\sqrt c }}} \right)^2} \le \frac{{27}}{{16}}\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) = \frac{{27}}{{16}}.\frac{{ab + bc + ca}}{{abc}} \le \frac{{81}}{{16}}.\frac{1}{{abc}} \end{array}$ Hence, $S \le \frac{3}{2}\sqrt[4]{{\frac{1}{{abc}}}}$. Done!

cjquines0 wrote: Let $a,b,c$ be positive real numbers with $a+b+c=3$. Prove that \[ \sqrt{\frac{b}{a^2+3}}+ \sqrt{\frac{c}{b^2+3}}+ \sqrt{\frac{a}{c^2+3}} ~\le~ \frac32\sqrt[4]{\frac{1}{abc}}\]

My solution.

cjquines0 wrote: Let $a,b,c$ be positive real numbers with $a+b+c=3$. Prove that \[ \sqrt{\frac{b}{a^2+3}}+ \sqrt{\frac{c}{b^2+3}}+ \sqrt{\frac{a}{c^2+3}} ~\le~ \frac32\sqrt[4]{\frac{1}{abc}}\] Let $a,b,c,d$ be positive real numbers with $a+b+c+d=4$. Prove that \[ \sqrt{\frac{b}{a^2+3}}+ \sqrt{\frac{c}{b^2+3}}+\sqrt{\frac{d}{c^2+3}} +\sqrt{\frac{a}{d^2+3}} \le 2\sqrt[4]{\frac{1}{abcd}}.\]Let $a_1,a_2,\cdots,a_n (n\ge 3)$ be positive real numbers such that $a_1+a_2+\cdots+a_n=n.$ Prove that $$\sqrt{\frac{a_2}{a^2_1+3}}+\sqrt{\frac{a_3}{a^2_2+3}}+\cdots+\sqrt{\frac{a_n}{a^2_{n-1}+3}}+\sqrt{\frac{a_1}{a^2_n+3}}\leq \frac{n}{2}\sqrt[4]{\frac{1}{a_1a_2\cdots a_n}}.$$

The inequality is equivalent to showing that $\sum\nolimits \sqrt[4]{\frac{16ab^3c}{(a^2 + 3)^2}} \le 3$, however $a^2 + 3 = a^2 + 1 + 1 + 1 \ge 4 \sqrt{a}$, therefore it is sufficient to show that $\sum\nolimits \sqrt[4]{\frac{16ab^3c}{16a}} \le 3$, i.e. $\sum\nolimits \sqrt[4]{b^3c} \le 3$, which follows directly from the $AM- GM$ inequality ($\sum\nolimits \sqrt[4]{b^3c} \le \sum\nolimits \frac{3b + c}{4} = \frac {4 \sum\nolimits a}{4} = \sum\nolimits a = 3$)