Let $ S_{n} = \frac{(3a)^n}{(b+1)(c+1)}+\frac{(3b)^n}{(c+1)(a+1)}+\frac{(3c)^n}{(a+1)(b+1)}$ By Chebyshev's inequality: $ S_{n+1} \geq \frac{1}{3}(3a+3b+3c)S_{n}=S_{n}$ so $ S_{n} \geq S_{0}=\frac{4}{(a+1)(b+1)(c+1)}$ then:$ S_{n} \geq \frac{4}{(a+1)(b+1)(c+1)} \geq \frac{27*4}{(a+b+c+3)^3}=\frac{27}{16}$

We have this by Hölders inequality : $\swarrow$ $ \boxed{((a+1)(c+1)+(c+1)(b+1)+(a+1)(b+1)).LHS.(1+1+1)^{n-2}\ge (3b+3a+3c)^n} $. Therefore, $LHS\ge \frac{9}{ab+ac+bc}$. Then we are done since $ab+ac+bc\le \frac{1}{3}$.

By holders $(LHS)(\cyc{b+1 + c+1 + a+1)(\cyc{c+1 + a+1 + b+1){(1 + 1 + 1)}^{n-3} \ge {(3a + 3b + 3c)}^n$ $16*3^{n-3}(LHS) \ge 3^n$ $LHS \ge 27/16$

WLOG assume that $a \ge b \ge c$. Then, $(3a)^{n-1} \ge (3b)^{n-1} \ge (3c)^{n-1}$. Also, $(b+1)(c+1) \le (c+1)(a+1) \le (a+1)(b+1)$. Using Chebyshev's Inequality, $\sum_{cyc} \frac{(3a)^n}{(b+1)(c+1)} \ge \frac{(\sum (3a)^{n-1})(\sum \frac{3a}{(b+1)(c+1)})}{3}$. By Power Mean Inequality, $\sum (3a)^{n-1} \ge 3 (\frac{\sum (3a)}{3})^{n-1} =3$. By Titu's Lemma $\sum \frac{3a}{(b+1)(c+1)} \ge 3\frac{(a+b+c)^2}{3abc+2ab+2bc+2ca+a+b+c} \ge \frac{27}{16}$ using $abc \le (\frac{a+b+c}{3})^3$ and $ab+bc+ca \le \frac{(a+b+c)^2}{3}$. Multiplying we get $\sum \frac{(3a)^n}{(b+1)(c+1)} \ge \frac{3.\frac{27}{16}}{3}=\frac{27}{16}$.

Goutham wrote: Let $a, b, c$ be positive real numbers such that $a + b + c = 1$. If $n$ is a positive integer then prove that \[ \frac{(3a)^n}{(b + 1)(c + 1)} + \frac{(3b)^n}{(c + 1)(a + 1)} + \frac{(3c)^n}{(a + 1)(b + 1)} \ge \frac{27}{16} \,. \] Really simple man . By holder inequality $(\frac{(3a)^n}{(b + 1)(c + 1)} + \frac{(3b)^n}{(c + 1)(a + 1)} + \frac{(3c)^n}{(a + 1)(b + 1)})(b+1+c+1+a+1)^2(1+1+1)^{n-3} \ge (3a+3b+3c)^n $ Thus it remains to prove that , $ab+bc+ac \le \frac{1}{3}$ which is obvious . One could have thought of induction too . Base case : We have to prove that $\frac{a}{(b + 1)(c + 1)} + \frac{b}{(c + 1)(a + 1)} + \frac{c}{(a + 1)(b + 1)} \ge \frac{9}{16}$ Which is simple by holder inequality . Now after this , we proceed with our next step We multiply by $3(a+b+c)$ , use rearrangement inequality $2$ times and get the desired result .

WLOG $a \ge b \ge c$ then clearly the sequences $3a$,$3b$,$3c$ and $\frac{(3a)^{n-1}}{(b+1)(c+1)}$,.... are similarly sorted,to be more specific they both are increasing.We denote the term of the problem inL.H.S in the problem as $S_n$.Clearly in the base case $n=0$ we have to prove $S_0 = \frac{1}{(b+1)(c+1)}+... \ge \frac{27}{16}$. After some manipulation it is equivalent to $(a+1)(b+1)(c+1) \le (\frac{4}{3})^3$ which is just easy from AM-GM. Now let $S_{n-1} \ge \frac{27}{16}$ Now $S_n=(3a)(\frac{(3a)^{n-1}}{(b+1)(c+1)}+.... )\ge (3a+3b+3c)(\frac{S_{n-1}}{3})=S_{n-1}$(from Chebyshev) Wow,$S_n$ is an increasing sequence for $n$.Now it is obvious that if $S_{n-1} \ge \frac{27}{16}$ then so is $S_n$.We are done.Actually the problem can be made stronger by replacing $n$ as a non-negative integer in place of the said positive integer.

Goutham wrote: Let $a, b, c$ be positive real numbers such that $a + b + c = 1$. If $n$ is a positive integer then prove that \[ \frac{(3a)^n}{(b + 1)(c + 1)} + \frac{(3b)^n}{(c + 1)(a + 1)} + \frac{(3c)^n}{(a + 1)(b + 1)} \ge \frac{27}{16} \,. \] By Chebyshov $\sum_{cyc}\frac{(3a)^n}{(b + 1)(c + 1)}\geq\frac{1}{3}\sum_{cyc}(3a)^n\sum_{cyc}\frac{1}{(1+b)(1+c)}=$ $=\frac{4\sum\limits_{cyc}(3a)^n}{3(1+a)(1+b)(1+c)}\geq\frac{4\cdot3}{3(1+a)(1+b)(1+c)}=\frac{4}{(1+a)(1+b)(1+c)}$. Thus, it remains to prove that $64(a+b+c)^3\geq27(2a+b+c)(2b+a+c)(2c+a+b)$, which is AM-GM.

Since $f(x)=x^n$ is convex for natural n, so by jensen $\sum a^n \ge 3 ({\frac{a+b+c}{3}})^n=\frac{1}{3^{n-1}}$ Also for a,b,c positive we have, $\sum \frac{1}{(a+1)(b+1)}= \frac{4}{(a+1)(b+1)(c+1)} \ge \frac{4}{({\frac{a+b+c+3}{3})}^3} =\frac{27}{16}$ wlog $a \ge b \ge c$ Then $(3a)^n \ge (3b)^n \ge (3c)^n$ $\frac{1}{(b+1)(c+1)} \ge \frac{1}{(a+1)(c+1)} \ge \frac{1}{(a+1)(b+1)}$ So by Chebyshev. $\sum \frac{(3a)^n}{(b+1)(c+1)} \ge 3^n\frac{\sum a^n}{3}\sum{\frac{1}{(a+1)(b+1)}} \ge 3^n \frac{1}{3^n}\frac{27}{16}=\frac{27}{16}$

Goutham wrote: Let $a, b, c$ be positive real numbers such that $a + b + c = 1$. If $n$ is a positive integer then prove that \[ \frac{(3a)^n}{(b + 1)(c + 1)} + \frac{(3b)^n}{(c + 1)(a + 1)} + \frac{(3c)^n}{(a + 1)(b + 1)} \ge \frac{27}{16} \,. \] Generalization of Yangzhiming： Let $x_1,x_2,\cdots,x_m$ $(m\geq 3)$ be positive real numbers such that $x_1+x_2+\cdots+x_m= 1$. If $k\geq0 $ then prove that

The first thing that comes to mind actually kills this...

Goutham wrote: Let $a, b, c$ be positive real numbers such that $a + b + c = 1$. If $n$ is a positive integer then prove that \[ \frac{(3a)^n}{(b + 1)(c + 1)} + \frac{(3b)^n}{(c + 1)(a + 1)} + \frac{(3c)^n}{(a + 1)(b + 1)} \ge \frac{27}{16} \,. \] n=1 wrote: The inequality crumbles to $16\sum a^2 \geq 9abc+9\sum ab + 2$.But since \begin{align*}9\sum a^2\geq 9\sum ab \\ abc \leq \dfrac{(a+b+c)^3}{27}=\dfrac{1}{27} \\ \sum a^2\geq \dfrac{(a+b+c)^2}{3}=\dfrac{1}{3}\end{align*}So this case follows.$\square$ n=2 wrote: Since $\sum ab\leq \dfrac{1}{3}$\[\frac{(3a)^2}{(b + 1)(c + 1)} + \frac{(3b)^2}{(c + 1)(a + 1)} + \frac{(3c)^2}{(a + 1)(b + 1)}\geq \frac{(3(a+b+c))^2}{\sum ab +5}\geq \dfrac{9}{\left(\dfrac{1}{3}+5\right)}=\dfrac{27}{16}\square\] n>=3 wrote: Let the expression be $T$.Then by Holders Inequality we have \[T\cdot ((b+1)+(c+1)+(a+1))\cdot ((c+1)+(a+1)+(b+1))\cdot \underbrace{(1+1+\cdots +1)\cdots (1+1+\cdots +1)}_{n-3 \; \text{times}}\geq (3(a+b+c))^n=3^n \newline \Longleftrightarrow T\geq \dfrac{27}{16}\square\]

very nice solution!

Pluto1708 wrote: Goutham wrote: Let $a, b, c$ be positive real numbers such that $a + b + c = 1$. If $n$ is a positive integer then prove that \[ \frac{(3a)^n}{(b + 1)(c + 1)} + \frac{(3b)^n}{(c + 1)(a + 1)} + \frac{(3c)^n}{(a + 1)(b + 1)} \ge \frac{27}{16} \,. \] n=1 wrote: The inequality crumbles to $16\sum a^2 \geq 9abc+9\sum ab + 2$.But since \begin{align*}9\sum a^2\geq 9\sum ab \\ abc \leq \dfrac{(a+b+c)^3}{27}=\dfrac{1}{27} \\ \sum a^2\geq \dfrac{(a+b+c)^2}{3}=\dfrac{1}{3}\end{align*}So this case follows.$\square$ n=2 wrote: Since $\sum ab\leq \dfrac{1}{3}$\[\frac{(3a)^2}{(b + 1)(c + 1)} + \frac{(3b)^2}{(c + 1)(a + 1)} + \frac{(3c)^2}{(a + 1)(b + 1)}\geq \frac{(3(a+b+c))^2}{\sum ab +5}\geq \dfrac{9}{\left(\dfrac{1}{3}+5\right)}=\dfrac{27}{16}\square\] n>=3 wrote: Let the expression be $T$.Then by Holders Inequality we have \[T\cdot ((b+1)+(c+1)+(a+1))\cdot ((c+1)+(a+1)+(b+1))\cdot \underbrace{(1+1+\cdots +1)\cdots (1+1+\cdots +1)}_{n-3 \; \text{times}}\geq (3(a+b+c))^n=3^n \newline \Longleftrightarrow T\geq \dfrac{27}{16}\square\] Nice! Same as my solution.

Goutham wrote: Let $a, b, c$ be positive real numbers such that $a + b + c = 1$. If $n$ is a positive integer then prove that \[ \frac{(3a)^n}{(b + 1)(c + 1)} + \frac{(3b)^n}{(c + 1)(a + 1)} + \frac{(3c)^n}{(a + 1)(b + 1)} \ge \frac{27}{16} \,. \] $1)n=1 \sum \frac{3a}{(b+1)(c+1)} \geq \frac{27}{16}$ $$16a^2+ 16b^2+16c^2+16 \geq 9(abc+ab+ac+bc+2)$$$$1=(a+b+c)^3 \geq 27abc \implies \frac{1}{3} \geq 9abc$$$$16a^2+ 16b^2+16c^2 \geq 9ab + 9 ac + 9bc + \frac{7}{3}$$$$a^2+b^2+c^2 \geq ab+ac+bc$$$$3(a^2+b^2+c^2) \geq (a+b+c)^2 =1$$can be proved by adding the above two. $2) n\geq 2$ $$(\frac{(3a)^n}{(b + 1)(c + 1)} + \frac{(3b)^n}{(c + 1)(a + 1)} + \frac{(3c)^n}{(a + 1)(b + 1)})(\sum(b+1)(c+1))(1+1+1)^{n-2} \ge (3a+3b+3c)^n $$remaining easy.$\blacksquare$