<s>[INST] 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
 [/INST] Even better we can compute an excellent approximation by the standard repertoire of functions of a complex variable to the value of this product, call it $P$. This has a certain didactic value. Introduce
$$S = \log P =\sum_{n\ge 1} \log\left(1-\frac{1}{3^n}\right).$$
The value $S$ is taken on at $x=1$ of the following harmonic sum:
$$S(x) = \sum_{n\ge 1} \log\left(1-\frac{1}{3^{xn}}\right).$$
Now we will evaluate this by inverting its Mellin transform.
Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = 1, \quad \mu_k = k \quad \text{and} \quad
g(x) = \log\left(1-\frac{1}{3^x}\right).$$
We need the Mellin transform $g^*(s)$ of $g(x)$ which is
$$\int_0^\infty \log\left(1-\frac{1}{3^x}\right) x^{s-1} dx
= \left[\log\left(1-\frac{1}{3^x}\right) \times \frac{x^s}{s} \right]_0^\infty
- \int_0^\infty \frac{\log 3}{3^x-1} \frac{x^s}{s} dx .$$
The first term in the difference vanishes for $\Re(s)\ge 1.$ It remains to do the calculation of the integral, which goes like this:
$$\int_0^\infty \frac{1}{3^x-1} x^s dx
= \int_0^\infty \frac{3^{-x}}{1-3^{-x}} x^s dx
= \int_0^\infty 3^{-x}\sum_{q\ge 0} 3^{-qx} x^s dx
= \sum_{q\ge 0} \int_0^\infty 3^{-(q+1)x} x^s dx.$$
Continuing, we find
$$\sum_{q\ge 0} \int_0^\infty e^{-(q+1)\log 3 \times x} x^s dx
= \sum_{q\ge 0} \frac{1}{(\log 3\times (q+1))^{s+1}}\Gamma(s+1)
= \frac{\Gamma(s+1)\zeta(s+1)}{(\log 3)^{s+1}}.$$
This gives that
$$g^*(s) = - \frac{\log 3}{s} \frac{\Gamma(s+1)\zeta(s+1)}{(\log 3)^{s+1}}
= - \frac{1}{s} \frac{\Gamma(s+1)\zeta(s+1)}{(\log 3)^s}.$$
By the harmonic sum identity we thus have for $Q(s)$, the Mellin transform of $S(x)$, that
$$Q(s) = - \frac{1}{s (\log 3)^s} \Gamma(s+1)\zeta(s)\zeta(s+1).$$
Now inverting the transform in the left half-plane to the left of $\Re(s)=3/2$ for an expansion about zero we find that the trivial zeros of the two zeta function terms cancel the poles of the gamma function, so that we are left with just three poles.
We get
$$\begin{align}
\mathrm{Res}(Q(s)/x^s; s=1) &= -\frac{1}{6} \frac{\pi^2}{\log 3} \frac{1}{x},\\
\mathrm{Res}(Q(s)/x^s; s=0) &= \frac{1}{2} \log \frac{2\pi}{\log 3} - \frac{1}{2}\log x,
\quad\text{and}\\
\mathrm{Res}(Q(s)/x^s; s=-1) &= \frac{1}{24} x \log 3.
\end{align}$$
Setting $x=1$ we obtain an approximation of $S(1)$ which is good to fifteen digits, namely
$$S(1) = S \approx 
-\frac{1}{6} \frac{\pi^2}{\log 3} + \frac{1}{2} \log \frac{2\pi}{\log 3} 
+ \frac{1}{24} \log 3\\
\approx -0.57959338143590694$$
whereas the exact value is $-0.57959338143590718.$
This finally gives the following approximate formula for $P$ (also good to fifteen digits): 
$$ P \approx \exp\left(-\frac{1}{6} \frac{\pi^2}{\log 3}\right) \times
\sqrt{\frac{2\pi}{\log 3} } \times 3^{1/24}
\approx 0.56012607792794900$$
whereas the exact value is $0.56012607792794894.$
It seems interesting to ask whether there is a better approximation using the standard repertoire than the one given above and it might be an instructive exercise (and produce a better approximation) to extract the first few terms of the sum $S(x)$ (using their exact values) and invert the Mellin transform of the rest.
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