How prove this inequality $\left(1-\frac{1}{3}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{3^n}\right)\ge\frac{14}{25}$ 
show that
$$\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3^2}\right)\cdots\left(1-\dfrac{1}{3^n}\right)\ge\dfrac{14}{25}\tag{1}$$

My try:
I only prove following  not strong inequality:

$$\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3^2}\right)\cdots\left(1-\dfrac{1}{3^n}\right)\ge\dfrac{1}{2}$$
proof: use Bernoulli inequality
$$(1+x_{1})(1+x_{2})\cdots (1+x_{n})\ge 1+x_{1}+x_{2}+\cdots +x_{n},$$where $x_{i}\ge -1$

so

$$\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3^2}\right)\cdots\left(1-\dfrac{1}{3^n}\right)\ge1-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{3^n}\right)$$
so
$$LHS\ge1-\sum_{n=1}^{\infty}\dfrac{1}{3^n}=1-\dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}=\dfrac{1}{2}$$

But for $(1)$,I can't prove it,Thank you

Here's another, somewhat complicated way to prove things.  We can rewrite the inequality (with $n$ taken to infinity) as
$${25\over14}\ge\prod_{k=1}^\infty{1\over1-(1/3)^k}=\sum_{k=0}^\infty p(k)(1/3)^k$$
where $p(k)$ is the partition function $1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56,77,101,\ldots$.  Let's accept for the moment the inequality
$$p(k)\le{1\over50}2^k\text{ for } k\gt11$$
Then we have
$$\sum_{k=0}^\infty p(k)(1/3)^k \le 1+{1\over3}+{2\over3^2}+{3\over3^3}+{5\over3^4}+{7\over3^5}+{11\over3^6}+{15\over3^7}+{22\over3^8}+{30\over3^9}+{42\over3^{10}}+{56\over3^{11}}+{3\over50}\left({2\over3}\right)^{12}$$
where the final term comes from
$$p(12)(1/3)^{12}+\cdots \le {1\over50}\left({2\over3}\right)^{12}\left(1+{2\over3}+\left({2\over3}\right)^2+\cdots\right)={1\over50}\left({2\over3}\right)^{12}{1\over1-(2/3)}={3\over50}\left({2\over3}\right)^{12}$$
The sum of all those fractions is not a lot of fun to work out, but it turns out to give 
$$\sum_{k=0}^\infty p(k)(1/3)^k \le {316217\over177147}+{3\over50}\left({2\over3}\right)^{12}\approx1.7855\lt1.7857\approx{25\over14}$$
The inequality on partition numbers that made this work can be proven by induction from Euler's pentagonal number theorem:
$$\begin{align}
p(k)&=p(k-1)+p(k-2)-p(k-5)-p(k-7)+p(k-12)+p(k-15)-\cdots\cr
&=p(k-1)+p(k-2)-\left(p(k-5)-p(k-12)\right)-\left(p(k-7)-p(k-15)\right)-\cdots\cr
&\le p(k-1)+p(k-2)\cr
\end{align}$$
Clearly one can conclude from this an inequality of the form $p(k)\le cr^k$ with any ratio $r$ greater than or equal to the golden ratio $\phi=(1+\sqrt5)/2$, but with a constant $c$ and/or starting value for $k$ that need to be checked against the initial terms of the sequence.  To the extent that I played around with different possibilities, I always found I had to go out to the twelfth term to make things less than $25/14$.
