It's a very known inequality.

We need to prove that $(a_{1}a_{2}\cdots a_{n})^{4} \ge (n - 1)^{n}$ We have, $\frac{a_{1}^{4}}{1 + a_{1}^{4}} = \frac{1}{1 + a_{2}^{4}} + \cdots + \frac{1}{1 + a_{n}^{4}}$, apply A - G, we get $\frac{a_{1}^{4}}{1 + a_{1}^{4}} \ge (n - 1)\sqrt[n - 1]{\frac{1}{(1 + a_{2}^{4})\cdots (1 + a_{n}^{4})}}$ Similarly, we also get $\frac{a_{k}^{4}}{1 + a_{k}^{4}} \ge (n - 1)\sqrt[n - 1]{\frac{1 + a_{k}^{4}}{(1 + a_{1}^{4})(1 + a_{2}^{4})\cdots (1 + a_{n}^{4})}}$ This implies $\frac{(a_{1}a_{2}\cdots a_{n})^{4}}{(1 + a_{1}^{4})(1 + a_{2})^{4}\cdots (1 + a_{n}^{4})} \ge (n - 1)^{n}\frac{1}{(1 + a_{1}^{4})(1 + a_{2}^{4})\cdots (1 + a_{n}^{4})}$ From this we implies our result.

LATVIA 2002

krenkovr wrote: LATVIA 2002 Let $ a,b,c,d>0$ and $ \dfrac{1}{1+a^4}+\dfrac{1}{1+b^4}+\dfrac{1}{1+c^4}+\dfrac{1}{1+d^4}=1$. Prove that:$$ abcd\ge 3$$

Inequalities: Theorems, Techniques and Selected Problems 2012

sqing wrote: Inequalities: Theorems, Techniques and Selected Problems 2012 We need to prove \[ \sum_{1}^{n} {\frac{1998}{x_i+1998}} = 1 \]\[ \sum_{1}^{n} {\frac{1}{\frac{x_i}{1998}+1}} = 1 \]Substituting $y_i = \frac{x_i}{1998}$ We have \[ \sum_{1}^{n} {\frac{1}{y_i+1}} = 1 \]and we need to prove \[ y_1y_2 \cdots y_n \geq (n-1)^n \]which is indeed very similar to the original post with $a_i^4 = y_i$

Stefan4024 wrote: Let $n\ge3$ and $a_1,a_2,...,a_n \in \mathbb{R^{+}}$, such that $\frac{1}{1+a_1^4} + \frac{1}{1+a_2^4} + ... + \frac{1}{1+a_n^4} = 1$. Prove that: $$a_1a_2...a_n \ge (n-1)^{\frac n4}$$

Let $a_1,a_2,\cdots,a_n $ be positive numbers such that $\frac{1}{a_1} + \frac{1}{a_2} + \cdots+ \frac{1}{a_n} = 1$$(n\geq 2 )$. Prove that: $$\sqrt[3]{\frac{a^3_1+a^3_2}{2}}+\sqrt[3]{\frac{a^3_2+a^3_3}{2}}+ \cdots+\sqrt[3]{\frac{a^3_{n-1}+a^3_n}{2}}+\sqrt[3]{\frac{a^3_n+a^3_1}{2}}+n\leq 2(a_1+a_2+\cdots+a_n).$$Let $a,b,c$ be non-negative numbers such that $a+b+c=3.$ For $n\geq 28 ,$ prove that $$\frac{1}{(a+b)^2+n}+\frac{1}{(b+c)^2+n}+\frac{1}{(c+a)^2+n}\leq \frac{3}{n+4}.$$(Marin Chirciu) CYCLIC-INEQUALITY-585、586

The following is also true: Let $a,b,c$ be nonnegative numbers such that $a+b+c=3.$ For $n\geq 14$, prove that $$\frac{1}{(a+b)^2+n}+\frac{1}{(b+c)^2+n}+\frac{1}{(c+a)^2+n}\leq \frac{3}{n+4}.$$

sqing wrote: Let $a_1,a_2,\cdots,a_n $ be positive numbers such that $\frac{1}{a_1} + \frac{1}{a_2} + \cdots+ \frac{1}{a_n} = 1$$(n\geq 2 )$. Prove that: $$\sqrt[3]{\frac{a^3_1+a^3_2}{2}}+\sqrt[3]{\frac{a^3_2+a^3_3}{2}}+ \cdots+\sqrt[3]{\frac{a^3_{n-1}+a^3_n}{2}}+\sqrt[3]{\frac{a^3_n+a^3_1}{2}}+n\leq 2(a_1+a_2+\cdots+a_n).$$ Use $\sqrt[3]{\frac{a^3+b^3}{2}}\le \frac{a^2+b^2}{a+b}=a+b-\frac{2ab}{a+b}=a+b-\frac{2}{\frac1a+\frac1b}$. Further use Cauchy-Schwarz Inequality.

MariusStanean wrote: The following is also true: Let $a,b,c$ be nonnegative numbers such that $a+b+c=3.$ For $n\geq 14$, prove that $$\frac{1}{(a+b)^2+n}+\frac{1}{(b+c)^2+n}+\frac{1}{(c+a)^2+n}\leq \frac{3}{n+4}.$$ Nice. Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be nonnegative real numbers such that $a_1+a_2+\cdots+a_n= n .$ Prove that $$ (1 + a_1)(1 + a_2)\cdots (1 + a_n) \leq (1 + a^2_1)(1 + a^2_2)\cdots (1 + a^2_n) .$$

sqing wrote: MariusStanean wrote: The following is also true: Let $a,b,c$ be nonnegative numbers such that $a+b+c=3.$ For $n\geq 14$, prove that $$\frac{1}{(a+b)^2+n}+\frac{1}{(b+c)^2+n}+\frac{1}{(c+a)^2+n}\leq \frac{3}{n+4}.$$ Nice. Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be nonnegative real numbers such that $a_1+a_2+\cdots+a_n= n .$ Prove that $$ (1 + a_1)(1 + a_2)\cdots (1 + a_n) \leq (1 + a^2_1)(1 + a^2_2)\cdots (1 + a^2_n) .$$ For real a,b,c is also true !

Stefan4024 wrote: Let $n\ge3$ and $a_1,a_2,...,a_n \in \mathbb{R^{+}}$, such that $\frac{1}{1+a_1^4} + \frac{1}{1+a_2^4} + ... + \frac{1}{1+a_n^4} = 1$. Prove that: $$a_1a_2...a_n \ge (n-1)^{\frac n4}$$ $$\iff$$Let $a_1,a_2,\cdots,a_n (n\ge 3)$ be positive real numbers such that $\frac{1}{1+a_1}+\frac{1}{1+a_2}+\cdots+\frac{1}{1+a_n}=1.$ Prove that $$a_1a_2\cdots a_n\geq(n-1)^n.$$