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This is true for the case $n = 2$, which we prove in the following form.

Main Result: $$P\Big[Y < Mean\Big] \ > \ \frac12$$

Proof: We change notation so that $X_i\sim LN (0, 2\sigma^2)$. Then $Mean=E[Y] = e^{\sigma^2}$.

Let \begin{align} U &= (\ln X_1 + \ln X_2)/2\\ V &= (\ln X_1 - \ln X_2)/2 \end{align} so that $U, V$ are iid $N (0, \sigma^2)$ with cdf $F$ and pdf $f$. Then: \begin{align} P\Big[Y<Mean\Big] &= P \left[\frac12(e^{U + V} + e^{U - V}) < e^{\sigma^2} \right] \\ &= P \left[e^U\cosh V < e^{\sigma^2} \right] \\ &= P \left[U < \sigma^{2^\phantom{2\!}} - \ln\cosh V \right] \\ &= \int_{-\infty}^\infty F\! \left[\sigma^2 - \ln\cosh v \right] f(v)\,dv \\ &> \int_{-\infty}^\infty \left (F[\sigma^2] + (1 - \cosh v)f(\sigma^2)\right) f(v)\,dv\\ &=F[\sigma^2]+\left(1-e^{\sigma^2/2}\right)f(\sigma^2)\\ &> 1/2 \end{align} The lemma below provides the first inequality here. The function of $\sigma$ which provides the second inequality is approximately $\frac12(1+\sigma/\sqrt{2\pi})$ for small $\sigma$ and $1-\sigma^{-1}/\sqrt{2\pi}$ for large $\sigma$.

Lemma: For $w>1$ (and in particular for $w=\cosh v$), $$F[\sigma^2 - \ln w] > F[\sigma^2] + (1-w)f(\sigma^2)$$ This is obviously true with equality when $w=1$. For $w>1$ it follows from the corresponding inequality for the derivatives of both sides, which can be written as: $$\frac{-f(\sigma^2 - \ln w)}{w} > -f(\sigma^2)$$ Cancelling many factors shows this equivalent to: $$\exp\left(\frac{-(\ln w)^2}{2\sigma^2}\right) < 1$$ which proves the lemma.

This is true for $n \le 4$, which we prove for $n=2$ in the following form, and then sketch for $n=3$ and $n=4$.

Main Result: $$P\Big[Y < Mean\Big] \ > \ \frac12$$

Proof: We change notation so that $X_i\sim LN (0, 2\sigma^2)$. Then $Mean=E[Y] = e^{\sigma^2}$.

Let \begin{align} U &= (\ln X_1 + \ln X_2)/2\\ V &= (\ln X_1 - \ln X_2)/2 \end{align} so that $U, V$ are iid $N (0, \sigma^2)$ with cdf $F$ and pdf $f$. Then: \begin{align} P\Big[Y<Mean\Big] &= P \left[\frac12(e^{U + V} + e^{U - V}) < e^{\sigma^2} \right] \\ &= P \left[e^U\cosh V < e^{\sigma^2} \right] \\ &= P \left[U < \sigma^{2^\phantom{2\!}} - \ln\cosh V \right] \\ &= E\! \left[F[\sigma^2 - \ln\cosh V]\right] \\ &> E\! \left[F[\sigma^2] - (\cosh V - 1)f(\sigma^2)\right]\\ &=F[\sigma^2]-\left(e^{\sigma^2/2}-1\right)f(\sigma^2)\\ &> 1/2 \end{align}

The first inequality here comes from the lemma below. 

The term $F[\sigma^2]$ in both inequalities is also $P[\sqrt{X_1 X_2}<Mean]$, and is approximately $\frac12+\sigma/\sqrt{2\pi}$ for small $\sigma$.

The function of $\sigma$ in the second inequality is approximately $\frac12(1+\sigma/\sqrt{2\pi})$ for small $\sigma$ and $1-\sigma^{-1}/\sqrt{2\pi}$ for large $\sigma$.

Lemma: For $W>1$ (and in particular for $W=\cosh V$), $$F[\sigma^2 - \ln W] > F[\sigma^2] - (W-1)f(\sigma^2)$$ This is obviously true with equality when $W=1$. For $W>1$ it follows from the corresponding inequality for the derivatives of both sides, which can be written as: $$\frac{-f(\sigma^2 - \ln W)}{W} > -f(\sigma^2)$$ Cancelling many factors shows this equivalent to: $$\exp\left(\frac{-(\ln W)^2}{2\sigma^2}\right) < 1$$ which proves the lemma.

Sketch of Extension for $2\le n\le 4$:

A similar technique proves the same result for $n \le 4$, transforming $\ln(X)$ by an orthogonal matrix where all entries in the first row are equal. Then we replace $e^U$ by the geometric mean of the $X$'s, and we replace $\cosh V$ by a function with $n$ exponential terms in $n-1$ variables. The corresponding lemma is: $$F[k\sigma^2 - k\ln W]\, > $$ $$F[k\sigma^2]\ -kf(k\sigma^2)\Big(W-1+\frac{k^2-1}{2}(W-1)^2\Big)$$ where $k=\sqrt{n/2}$; this lemma would be false for higher $n$ (e.g. at $n=7$, $\sigma=1$, $W=1.05$), but Bernoulli's inequality with an exponent of $0\le k^2-1 \le 1$ shows that it is true for $2\le n\le 4$. When we take expectations of this lemma, we get a good bound for low $\sigma$, which we can combine with separate and easier bounds for high $\sigma$ to prove the result.

This is true for the case $n = 2$.

We change notation so that $X_i\sim LN (0, 2\sigma^2)$. Then $Mean=E[Y] = e^{\sigma^2}$, and we are looking to show that $P\Big[Y<Mean\Big] > \frac12.$ We use $F$ and $f$ for the CDF and PDF of $N(0,\sigma^2)$ variables

Bound 1: $$P\Big[Y < Mean\Big] \ > \ P\Big[X < Mean\Big]^2 = \Phi\Big[\frac\sigma{\sqrt2}\Big]^2$$ The average is less than the mean if both variables are less than the mean, and this gives the desired inequality when $\sigma>0.78$.

Bound 2: When $\sigma < 2.96$, $$P\Big[Y < Mean\Big] \ > \ F\Big[\sigma^2\Big]-f(\sigma^2)\left(\frac{\sigma^2}{2}+\frac{\sigma^4}{8}\right).$$

The two bounds together establish the desired inequality $P\Big[Y<Mean\Big] > \frac12$.

Proof of Bound 2: Let \begin{align} U &= (\ln X_1 + \ln X_2)/2\\ V &= (\ln X_1 - \ln X_2)/2 \end{align} so that $U, V$ are iid $N (0, \sigma^2)$. Then: \begin{align} P\Big[Y<Mean\Big] &= P \left[\frac12(e^{U + V} + e^{U - V}) < e^{\sigma^2} \right] \\ &= P \left[e^U\cosh V < e^{\sigma^2} \right] \\ &= P \left[U < \sigma^{2^\phantom{2\!}} - \ln\cosh V \right] \\ &= \int_{-\infty}^\infty F\! \left[\sigma^2 - \ln\cosh v \right] f(v)\,dv \\ &\ge \int_{-\infty}^\infty \left (F[\sigma^2] - \frac{12 v^2 + v^4}{24} f(\sigma^2) \right) f(v)\,dv\\ &=F[\sigma^2]-f(\sigma^2)\left(\frac{\sigma^2}{2}+\frac{\sigma^4}{8}\right) \end{align} This last expression is always at least $1/2$, and is $\frac12 + \frac\sigma{2\sqrt{2\pi}} + O(\sigma^3)$ near $\sigma = 0$. The inequality is a comparison of a function with its fourth-order Taylor series, as in the following lemma.

Lemma: If $\sigma < 2.96$, $$F[\sigma^2 - \ln\cosh v] \ge F[\sigma^2] - \frac{12 v^2 + v^4}{24} f(\sigma^2)$$ This is obviously true with equality when $v=0$, and obviously symmetric in $v$. For positive $v$ it follows from the corresponding inequality for the derivatives of both sides, which can be written as: $$f(\sigma^2 - \ln\cosh v)(-\tanh v) \ge - \frac{6v + v^3}{6} f(\sigma^2)$$ This is equivalent to $$\sigma<\frac{\ln \cosh v}{\sqrt{2\ln(6 \sinh v/(6v+v^3))}}$$ and the last expression has a minimum value near $2.96$ (near $v=4.54$), proving the lemma.

The inequalities here are delicate: the lemma would fail for $\sigma\ge3$ (e.g. at $v=5$), and would fail for $\sigma$ near $0$ if the quartic term were omitted. So it's surprising and helpful that the required inequality between $v$ and $\sigma$ boils down to a single-variable function of $v$.

This is true for the case $n = 2$, which we prove in the following form.

Main Result: $$P\Big[Y < Mean\Big] \ > \ \frac12$$

Proof: We change notation so that $X_i\sim LN (0, 2\sigma^2)$. Then $Mean=E[Y] = e^{\sigma^2}$.

Let \begin{align} U &= (\ln X_1 + \ln X_2)/2\\ V &= (\ln X_1 - \ln X_2)/2 \end{align} so that $U, V$ are iid $N (0, \sigma^2)$ with cdf $F$ and pdf $f$. Then: \begin{align} P\Big[Y<Mean\Big] &= P \left[\frac12(e^{U + V} + e^{U - V}) < e^{\sigma^2} \right] \\ &= P \left[e^U\cosh V < e^{\sigma^2} \right] \\ &= P \left[U < \sigma^{2^\phantom{2\!}} - \ln\cosh V \right] \\ &= \int_{-\infty}^\infty F\! \left[\sigma^2 - \ln\cosh v \right] f(v)\,dv \\ &> \int_{-\infty}^\infty \left (F[\sigma^2] + (1 - \cosh v)f(\sigma^2)\right) f(v)\,dv\\ &=F[\sigma^2]+\left(1-e^{\sigma^2/2}\right)f(\sigma^2)\\ &> 1/2 \end{align} The lemma below provides the first inequality here. The function of $\sigma$ which provides the second inequality is approximately $\frac12(1+\sigma/\sqrt{2\pi})$ for small $\sigma$ and $1-\sigma^{-1}/\sqrt{2\pi}$ for large $\sigma$.

Lemma: For $w>1$ (and in particular for $w=\cosh v$), $$F[\sigma^2 - \ln w] > F[\sigma^2] + (1-w)f(\sigma^2)$$ This is obviously true with equality when $w=1$. For $w>1$ it follows from the corresponding inequality for the derivatives of both sides, which can be written as: $$\frac{-f(\sigma^2 - \ln w)}{w} > -f(\sigma^2)$$ Cancelling many factors shows this equivalent to: $$\exp\left(\frac{-(\ln w)^2}{2\sigma^2}\right) < 1$$ which proves the lemma.

This is true for the case $n = 2$.

We change notation so that $X_i\sim LN (0, 2\sigma^2)$. Then $Mean=E[Y] = e^{\sigma^2}$, and we are looking to show that $P\Big[Y<Mean\Big] > \frac12.$ We use $F$ and $f$ for the CDF and PDF of $N(0,\sigma^2)$ variables

Bound 1: $$P\Big[Y < Mean\Big] \ > \ P\Big[X < Mean\Big]^2 = \Phi\Big[\frac\sigma{\sqrt2}\Big]^2$$ The average is less than the mean if both variables are less than the mean, and this gives the desired inequality when $\sigma>0.78$.

Bound 2: When $\sigma < 2.96$, $$P\Big[Y < Mean\Big] \ > \ F\Big[\sigma^2\Big]-f(\sigma^2)\left(\frac{\sigma^2}{2}+\frac{\sigma^4}{8}\right).$$

The two bounds together establish the desired inequality $P\Big[Y<Mean\Big] > \frac12$.

Proof of Bound 2: Let \begin{align} U &= (\ln X_1 + \ln X_2)/2\\ V &= (\ln X_1 - \ln X_2)/2 \end{align} so that $U, V$ are iid $N (0, \sigma^2)$. Then: \begin{align} P\Big[Y<Mean\Big] &= P \left[\frac12(e^{U + V} + e^{U - V}) < e^{\sigma^2} \right] \\ &= P \left[e^U\cosh V < e^{\sigma^2} \right] \\ &= P \left[U < \sigma^{2^\phantom{2\!}} - \ln\cosh V \right] \\ &= \int_{-\infty}^\infty F\! \left[\sigma^2 - \ln\cosh v \right] f(v)\,dv \\ &\ge \int_{-\infty}^\infty \left (F[\sigma^2] - \frac{12 v^2 + v^4}{24\sigma} f(\sigma^2) \right) f(v)\,dv\\ &=F[\sigma^2]-f(\sigma^2)\left(\frac{\sigma^2}{2}+\frac{\sigma^4}{8}\right) \end{align} This last expression is always at least $1/2$, and is $\frac12 + \frac\sigma{2\sqrt{2\pi}} + O(\sigma^3)$ near $\sigma = 0$. The inequality is a comparison of a function with its fourth-order Taylor series, as in the following lemma.

Lemma: If $\sigma < 2.96$, $$F[\sigma^2 - \ln\cosh v] \ge F[\sigma^2] - \frac{12 v^2 + v^4}{24} f(\sigma^2)$$ This is obviously true with equality when $v=0$, and obviously symmetric in $v$. For positive $v$ it follows from the corresponding inequality for the derivatives of both sides, which can be written as: $$f(\sigma^2 - \ln\cosh v)(-\tanh v) \ge - \frac{6v + v^3}{6} f(\sigma^2)$$ This is equivalent to $$\sigma<\frac{\ln \cosh v}{\sqrt{2\ln(6 \sinh v/(6v+v^3))}}$$ and the last expression has a minimum near $2.96$ (near $v=4.54$), proving the lemma. The inequality here is delicate, because the minimum would be zero if the quartic term were omitted.

This is true for the case $n = 2$.

We change notation so that $X_i\sim LN (0, 2\sigma^2)$. Then $Mean=E[Y] = e^{\sigma^2}$, and we are looking to show that $P\Big[Y<Mean\Big] > \frac12.$ We use $F$ and $f$ for the CDF and PDF of $N(0,\sigma^2)$ variables

Bound 1: $$P\Big[Y < Mean\Big] \ > \ P\Big[X < Mean\Big]^2 = \Phi\Big[\frac\sigma{\sqrt2}\Big]^2$$ The average is less than the mean if both variables are less than the mean, and this gives the desired inequality when $\sigma>0.78$.

Bound 2: When $\sigma < 2.96$, $$P\Big[Y < Mean\Big] \ > \ F\Big[\sigma^2\Big]-f(\sigma^2)\left(\frac{\sigma^2}{2}+\frac{\sigma^4}{8}\right).$$

The two bounds together establish the desired inequality $P\Big[Y<Mean\Big] > \frac12$.

Proof of Bound 2: Let \begin{align} U &= (\ln X_1 + \ln X_2)/2\\ V &= (\ln X_1 - \ln X_2)/2 \end{align} so that $U, V$ are iid $N (0, \sigma^2)$. Then: \begin{align} P\Big[Y<Mean\Big] &= P \left[\frac12(e^{U + V} + e^{U - V}) < e^{\sigma^2} \right] \\ &= P \left[e^U\cosh V < e^{\sigma^2} \right] \\ &= P \left[U < \sigma^{2^\phantom{2\!}} - \ln\cosh V \right] \\ &= \int_{-\infty}^\infty F\! \left[\sigma^2 - \ln\cosh v \right] f(v)\,dv \\ &\ge \int_{-\infty}^\infty \left (F[\sigma^2] - \frac{12 v^2 + v^4}{24} f(\sigma^2) \right) f(v)\,dv\\ &=F[\sigma^2]-f(\sigma^2)\left(\frac{\sigma^2}{2}+\frac{\sigma^4}{8}\right) \end{align} This last expression is always at least $1/2$, and is $\frac12 + \frac\sigma{2\sqrt{2\pi}} + O(\sigma^3)$ near $\sigma = 0$. The inequality is a comparison of a function with its fourth-order Taylor series, as in the following lemma.

Lemma: If $\sigma < 2.96$, $$F[\sigma^2 - \ln\cosh v] \ge F[\sigma^2] - \frac{12 v^2 + v^4}{24} f(\sigma^2)$$ This is obviously true with equality when $v=0$, and obviously symmetric in $v$. For positive $v$ it follows from the corresponding inequality for the derivatives of both sides, which can be written as: $$f(\sigma^2 - \ln\cosh v)(-\tanh v) \ge - \frac{6v + v^3}{6} f(\sigma^2)$$ This is equivalent to $$\sigma<\frac{\ln \cosh v}{\sqrt{2\ln(6 \sinh v/(6v+v^3))}}$$ and the last expression has a minimum value near $2.96$ (near $v=4.54$), proving the lemma.

The inequalities here are delicate: the lemma would fail for $\sigma\ge3$ (e.g. at $v=5$), and would fail for $\sigma$ near $0$ if the quartic term were omitted. So it's surprising and helpful that the required inequality between $v$ and $\sigma$ boils down to a single-variable function of $v$.