(14) Given the function $f(x)=x^{2}+x+\sqrt{3}$, if for all positive numbers $a, b, c$, the inequality $f\left(\frac{a+b+c}{3}-\sqrt[3]{a b c}\right) \geqslant f\left(\lambda\left(\frac{a+b}{2}-\sqrt{a b}\right)\right)$ always holds, find the maximum value of the positive number $\lambda$.
14 Because $f(x)=x^{2}+x+\sqrt{3}$ is monotonically increasing on the interval $\left(-\frac{1}{2},+\infty\right)$, and when $a, b, c, \lambda$ are all positive numbers,
$$
\begin{array}{c}
\frac{a+b+c}{3}-\sqrt[3]{a b c}>0, \\
\frac{a+b}{2}-\sqrt{a b}>0,
\end{array}
$$

so
$$
f\left(\frac{a+b+c}{3}-\sqrt[3]{a b c}\right) \geqslant f\left(\lambda\left(\frac{a+b}{2}-\sqrt{a b}\right)\right),
$$

is equivalent to
$$
\frac{a+b+c}{3}-\sqrt[3]{a b c} \geqslant \lambda\left(\frac{a+b}{2}-\sqrt{a b}\right),
$$

then the original problem is transformed into
$$
\frac{\frac{a+b+c}{3}-\sqrt[3]{a b c}}{\frac{a+b}{2}-\sqrt{a b}} \geqslant \lambda
$$

for all positive numbers $a, b, c$. Find the maximum value of the positive number $\lambda$.
When $c=\mu^{3} \sqrt{a b}(\mu>0)$,
$$
\begin{aligned}
\frac{\frac{a+b+c}{3}-\sqrt[3]{a b c}}{\frac{a+b}{2}-\sqrt{a b}} & =\frac{\frac{a+b+\mu^{3} \sqrt{a b}}{3}-\sqrt[3]{a b\left(\mu^{3} \sqrt{a b}\right)}}{\frac{a+b}{2}-\sqrt{a b}} \\
& =\frac{2}{3} \cdot \frac{a+b+\left(\mu^{3}-3 \mu\right) \sqrt{a b}}{a+b-2 \sqrt{a b}} \\
& \geqslant \lambda .
\end{aligned}
$$

Let $\mu^{3}-3 \mu=-2$, then $\mu=1$, at this time $\lambda \leqslant \frac{2}{3}$ holds. According to this, we guess $\lambda_{\text {max }}=\frac{2}{3}$, so we only need to prove: for all positive numbers $a, b, c$,
$$
\frac{\frac{a+b+c}{3}-\sqrt[3]{a b c}}{\frac{a+b}{2}-\sqrt{a b}} \geqslant \frac{2}{3} .
$$

And
$$
\frac{\frac{a+b+c}{3}-\sqrt[3]{a b c}}{\frac{a+b}{2}-\sqrt{a b}} \geqslant \frac{2}{3},
$$

is equivalent to
$$
3\left(\frac{a+b+c}{3}-\sqrt[3]{a b c}\right) \geqslant 2\left(\frac{a+b}{2}-\sqrt{a b}\right),
$$

which is equivalent to
$$
c+2 \sqrt{a b} \geqslant 3 \sqrt[3]{a b c} .
$$

Since
$$
c+2 \sqrt{a b}=c+\sqrt{a b}+\sqrt{a b} \geqslant 3 \sqrt[3]{a b c},
$$

for all positive numbers $a, b, c$,
$$
\frac{\frac{a+b+c}{3}-\sqrt[3]{a b c}}{\frac{a+b}{2}-\sqrt{a b}} \geqslant \frac{2}{3} .
$$

Therefore, the maximum value of $\lambda$ is $\frac{2}{3}$.