parmenides51 wrote: Prove that all postive real numbers $a,b,c$ with $a+b+c=4$ satisfy the inequality $$\frac{ab}{\sqrt[4]{3c^2+16}}+ \frac{bc}{\sqrt[4]{3a^2+16}}+ \frac{ca}{\sqrt[4]{3b^2+16}} \le\frac43 \sqrt[4]{12}$$ $$\sum_{cyc}\frac{ab}{\sqrt[4]{3c^2+16}}\leq \frac{\sqrt 2}{\sqrt[4] 3}\sum_{cyc}\frac{ab}{\sqrt{c+4}}\leq \frac{\sqrt 2}{\sqrt[4]3}\sqrt{\sum_{cyc} ab\sum_{cyc}\frac{ab}{c+4}}\leq \frac{4\sqrt 2}{\sqrt[4]{27}}\sqrt{\sum_{cyc}\frac{ab}{(b+c)+(c+a)}}\leq\frac{4\sqrt[4]{12}}{3 }\sqrt{\frac{1}{4}\sum_{cyc}\left(\frac{ab}{b+c}+\frac{ab}{c+a}\right)}=\frac43 \sqrt[4]{12}.$$ Consider real numbers $a, b ,c \ge0$ with $a+b+c=2$. Prove that: $$\frac{ab}{\sqrt[4]{3c^2+4}} +\frac{bc}{\sqrt[4]{3a^2+4}}+\frac{ca}{\sqrt[4]{3b^2+4}}\le \frac{2\sqrt[4] {3}}{3}$$

In what follows below, we keep in mind that for $a=b=c=\frac43$, one enjoys the equality. We first get rid of $\sqrt[4]{12}$ as follows: it is equivalent to proving \[ \sum_{{\rm sym}}\frac{ab}{\sqrt[4]{36c^2+192}} \le \frac43. \]Next, we study the denominator. Note that \begin{align*} 36c^2+192 &= 36c^2+64+64+64\\ &=36c^2+4(a+b+c)^2+4(a+b+c)^2+4(a+b+c)^2. \end{align*}The idea behind this decomposition is to ensure that all four summands, in the equality case, are indeed equal (as they should be). Applying Cauchy-Schwarz, \[ 4\left(36c^2+4(a+b+c)^2+4(a+b+c)^2+4(a+b+c)^2\right)\ge (6c+24)^2. \]Consequently, \[ \frac{ab}{\sqrt[4]{36c^2+192}} \le \frac{ab}{\sqrt{3c+12}}. \]Hence, it suffices to show \[ \sum_{{\rm sym}} \frac{ab}{\sqrt{3c+12}} \le \frac43. \]Now, we apply Cauchy-Schwarz inequality to obtain \[ \left(\sum_{{\rm sym}} \frac{ab}{\sqrt{3c+12}} \right)^2 \le (ab+bc+ca)\left(\sum_{{\rm sym}} \frac{ab}{3c+12}\right). \]Since \[ 16=(a+b+c)^2 \ge 3(ab+bc+ca)\Rightarrow \frac{16}{3} \ge (ab+bc+ca), \]we have \[ (ab+bc+ca)\left(\sum_{{\rm sym}} \frac{ab}{3c+12}\right) \le \frac{16}{3}\left(\sum_{{\rm sym}} \frac{ab}{3c+12}\right). \]Thus to conclude, it suffices to show \[ \sum_{{\rm sym}}\frac{ab}{c+4}\le 1. \]To show this, we will use the simple inequality $\frac1x+\frac1y\ge \frac{4}{x+y}$. In particular, taking $x=a+c$ and $y=b+c$, we have \[ \frac{ab}{c+4}=\frac{ab}{(a+c)+(b+c)} \le \frac14 \left(\frac{ab}{a+c}+\frac{ab}{b+c}\right) \]But since \[ \sum \left(\frac{ab}{a+c}+\frac{ab}{b+c}\right) = a+b+c=4, \]we have the desired conclusion.

sqing wrote: Consider real numbers $a, b ,c \ge0$ with $a+b+c=2$. Prove that: $$\frac{ab}{\sqrt[4]{3c^2+4}} +\frac{bc}{\sqrt[4]{3a^2+4}}+\frac{ca}{\sqrt[4]{3b^2+4}}\le \frac{2\sqrt[4] {3}}{3}$$ $$\sum_{cyc}\frac{bc}{\sqrt[4]{9a^2+12}}\leq \sum_{cyc}\frac{bc}{\sqrt{\frac{3}{2}(a+2)}}\leq \sqrt{\frac{2}{3}\sum_{cyc} bc\sum_{cyc}\frac{bc}{a+2}}\leq\frac{2 \sqrt{2}}{3} \sqrt{\sum_{cyc}\frac{bc}{(c+a)+(a+b)}}\leq\frac{\sqrt{2}}{3} \sqrt{\sum_{cyc}\left(\frac{bc}{c+a}+\frac{bc}{a+b}\right)}=\frac{2}{3}.$$