crazyfehmy wrote: Prove that for all $n \in \mathbb{Z^+}$ and for all positive real numbers satisfying $a_1a_2...a_n=1$ \[ \displaystyle\sum_{i=1}^{n} \frac{a_i}{\sqrt{{a_i}^4+3}} \leq \frac{1}{2}\displaystyle\sum_{i=1}^{n} \frac{1}{a_i} \] See here.

$\textcolor{red}{Claim:}$ $\frac{a_i}{\sqrt{{a_i}^{4}+3}}\leq \frac{a_i}{a_i+1}$ for $a_i \in \mathbb{R^+}$ $\textcolor{red}{Proof:}$ $\frac{a_i}{\sqrt{{a_i}^{4}+3}}\leq \frac{a_i}{a_i+1} \Leftrightarrow (a_i+1)^{2}\leq({a_i}^{4}+3),$ meaning ${a_i}^{4}-{a_i}^2-2{a_i}+2=(a_i-1)^2((a_i+1)^{2}+1)\geq 0$ which is always right, thus our claim is right. Since $\displaystyle\sum_{i=1}^{n}\frac{a_i}{\sqrt{{a_i}^{4}+3}}\leq \displaystyle\sum_{i=1}^{n} \frac{a_i}{a_i+1}$ we just need to prove that $\displaystyle\sum_{i=1}^{n} \frac{a_i}{a_i+1}\leq \frac{1}{2} \displaystyle\sum_{i=1}^{n} \frac{1}{a_i}$ Let $f(x)=\frac{x}{x+1}.$ Since $f$ is a strictly increasing concave function over $\mathbb{R^+},$ from Jensen and AM-GM we have ${\displaystyle\sum_{i=1}^{n} \frac{a_i}{a_i+1}}\leq nf(1)=\frac{n}{2} \leq nf\left(\frac{\left(\displaystyle\sum_{i=1}^{n} {a_i}\right)}{n}\right).$ Lastly from AM-GM we have $n\leq \displaystyle\sum_{i=1}^{n} \frac{1}{a_i}$ meaning $\displaystyle\sum_{i=1}^{n}\frac{a_i}{\sqrt{{a_i}^{4}+3}}\leq \displaystyle\sum_{i=1}^{n} \frac{a_i}{a_i+1}\leq \frac{n}{2}\leq \frac{1}{2}\displaystyle\sum_{i=1}^{n} \frac{1}{a_i}$ and so we are done!

Captain_Baran wrote: $\sum \frac{a_i}{a_i+1}\leq \frac{n}{2}$ Maybe there is a mistake here since taking $a_1=a_2=a_3=5,a_4=\frac{1}{625}$ gives that $LHS>\frac{5}{2}>2=RHS$ for $n=4$. Taking $a_i$ to positive infinity for $i\geq 2$ and $a_1$ to $0$ gives a contradiction.