Having said that, I also believe that $\max(T,V) \ \big| \ V \leq T$
should have the same distribution as the random variable, $T + \min(T,V)$ which is apparently wrong.
Of course it's wrong: if you know that $\ V \leq T$, then you know that the maximum is $T$, hence $\max(T,V) \ \big| \ V \leq T$ has the same distribution as $T$.
Edited:
"Of course", the above it's totally wrong. Sorry. Knowing $\ V \leq T$ not only informs us that the maximum is $T$, but also tell us something about the value of the maximum (and it should push it expected value up ).
I think you are confusing two kind of knowledge (condition): which is the minimum, and what is the value of the minimum. If you know that the mininum is the variable $V$, and that its value is $v$ (don't confuse the random variables with their values), then you know that $T\ge v$, and in that case the distribution of the conditioned variable shifts by $v$.
WLOG, we assume that $\sigma>0$ and $\tau >0$. Then
\begin{align}
E(\max(X, Y)) &= \iint\limits_{\mu+\sigma z_1 \ge \nu + \tau z_2}e^{\mu+\sigma z_1}\frac{1}{2\pi}e^{-\frac{1}{2}(z_1^2+z_2^2)}dz_1 dz_2 \\
&\quad + \iint\limits_{\mu+\sigma z_1 \le \nu + \tau z_2}e^{\nu+\tau z_2}\frac{1}{2\pi}e^{-\frac{1}{2}(z_1^2+z_2^2)}dz_1 dz_2.
\end{align}
Making the substitutions
\begin{align*}
x = \frac{\sigma z_1 - \tau z_2}{\sqrt{\sigma^2+\tau^2}} \mbox{ and } y = \frac{\tau z_1 + \sigma z_2}{\sqrt{\sigma^2+\tau^2}},
\end{align*}
we obtain that
\begin{align*}
& \ \iint\limits_{\mu+\sigma z_1 \ge \nu + \tau z_2}e^{\mu+\sigma z_1}\frac{1}{2\pi}e^{-\frac{1}{2}(z_1^2+z_2^2)}dz_1 dz_2\\
=&\ \int_{-\infty}^{\infty} dy \int_{\frac{\nu-\mu}{\sqrt{\sigma^2+\tau^2}}}^{\infty}e^{\mu+\sigma \frac{\sigma x + \tau y}{\sqrt{\sigma^2+\tau^2}}}\frac{1}{2\pi}e^{-\frac{1}{2}(x^2+y^2)}dxdy\\
=&\ \int_{-\infty}^{\infty} dy \int_{\frac{\nu-\mu}{\sqrt{\sigma^2+\tau^2}}}^{\infty}e^{\mu+\frac{1}{2}\sigma^2}\frac{1}{2\pi}e^{-\frac{1}{2}\left[\big(x-\frac{\sigma^2}{\sqrt{\sigma^2+\tau^2}}\big)^2+\big(y-\frac{\sigma\tau}{\sqrt{\sigma^2+\tau^2}}\big)^2\right]}dxdy\\
=&\ \int_{\frac{\nu-\mu}{\sqrt{\sigma^2+\tau^2}}}^{\infty}e^{\mu+\frac{1}{2}\sigma^2}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\big(x-\frac{\sigma^2}{\sqrt{\sigma^2+\tau^2}}\big)^2}dx\\
=&\ \int_{-\infty}^{-\frac{\nu-\mu}{\sqrt{\sigma^2+\tau^2}}}e^{\mu+\frac{1}{2}\sigma^2}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\big(x+\frac{\sigma^2}{\sqrt{\sigma^2+\tau^2}}\big)^2}dx\\
=&\ e^{\mu+\frac{1}{2}\sigma^2} N\left(\frac{-\nu+\mu+\sigma^2}{\sqrt{\sigma^2+\tau^2}} \right),
\end{align*}
where $N$ is the cumulative distribution function of a standard normal random variable. Similarly,
\begin{align*}
\iint\limits_{\nu + \tau z_2 \ge \mu+\sigma z_1}e^{\nu+\tau z_2}\frac{1}{2\pi}e^{-\frac{1}{2}(z_1^2+z_2^2)}dz_1 dz_2 &= e^{\nu+\frac{1}{2}\tau^2} N\left(\frac{-\mu+\nu+\tau^2}{\sqrt{\sigma^2+\tau^2}} \right).
\end{align*}
That is,
\begin{align*}
E(\max(X, Y)) &= e^{\nu+\frac{1}{2}\tau^2} N\left(\frac{-\mu+\nu+\tau^2}{\sqrt{\sigma^2+\tau^2}} \right) + e^{\mu+\frac{1}{2}\sigma^2} N\left(\frac{-\nu+\mu+\sigma^2}{\sqrt{\sigma^2+\tau^2}} \right).
\end{align*}
Alternative solution.
Note that
\begin{align*}
\max(X, Y) &= Y + \max(X-Y, \ 0)\\
&= Y + Y\max\Big(\frac{X}{Y}-1, \ 0\Big).
\end{align*}
Let $P$ be the given probability measure, and $Q$ be a probability measure that is defined by the Randon-Nykodim derivative
\begin{align*}
\frac{dQ}{dP} = e^{-\frac{1}{2}\tau^2 + \tau Z_2}.
\end{align*}
Then
\begin{align*}
\widehat{Z}_1 = Z_1, \mbox{ and }
\widehat{Z}_2 = Z_2 - \tau
\end{align*}
are two standard independent normal random variables under the probability measure $Q$.
Moreover,
\begin{align*}
Z \equiv \frac{X}{Y} &= e^{\mu-\nu - \tau^2 + \sigma \widehat{Z}_1 - \tau \widehat{Z}_2}\\
&= e^{\mu-\nu - \tau^2 + \sqrt{\sigma^2+\tau^2} \xi},
\end{align*}
where
\begin{align*}
\xi = \frac{\sigma \widehat{Z}_1 - \tau \widehat{Z}_2}{\sqrt{\sigma^2+\tau^2}}
\end{align*}
is a standard normal random variable. Let
\begin{align*}
\lambda = \frac{-\mu+ \nu + \tau^2}{\sqrt{\sigma^2+\tau^2}},
\end{align*}
and $E_Q$ be the expectation operator under the measure $Q$.
Then,
\begin{align*}
E\big(\max(X, Y) \big) &= E(Y)+E\big(Y\max(Z-1, \ 0) \big)\\
&=E(Y)+e^{\nu+\frac{1}{2}\tau^2} E_Q\big(\max(Z-1, \ 0) \big).
\end{align*}
Furthermore,
\begin{align*}
E_Q\big(\max(Z-1, \ 0) \big) &= E_Q\left(Z\,\mathbb{I}_{Z \ge 1}\right) - E_Q\left(\mathbb{I}_{Z \ge 1}\right)\\
&=\int_{\lambda}^{\infty} \frac{1}{\sqrt{2\pi}}e^{\mu-\nu -\tau^2 + \sqrt{\sigma^2+\tau^2} z -\frac{1}{2}z^2} dz - N(-\lambda)\\
&=e^{\mu-\nu + \frac{1}{2} (\sigma^2-\tau^2)} \int_{\lambda}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\big(z-\sqrt{\sigma^2+\tau^2}\big)^2} dz - N(-\lambda)\\
&=e^{\mu-\nu + \frac{1}{2} (\sigma^2-\tau^2)}N\Big(\sqrt{\sigma^2+\tau^2} -\lambda\Big) - N(-\lambda).
\end{align*}
That is,
\begin{align*}
E\big(\max(X, Y) \big) &= e^{\nu + \frac{1}{2}\tau^2} \bigg[1 + e^{\mu-\nu + \frac{1}{2} (\sigma^2-\tau^2)}N\Big(\sqrt{\sigma^2+\tau^2} -\lambda\Big) - N(-\lambda) \bigg]\\
&=e^{\nu + \frac{1}{2}\tau^2} N(\lambda) + e^{\mu + \frac{1}{2} \sigma^2}N\Big(\sqrt{\sigma^2+\tau^2} -\lambda\Big)\\
&= e^{\nu + \frac{1}{2}\tau^2}N\left(\frac{-\mu+ \nu + \tau^2}{\sqrt{\sigma^2+\tau^2}}\right) + e^{\mu + \frac{1}{2} \sigma^2}N\left(\frac{-\nu+\mu+\sigma^2}{\sqrt{\sigma^2+\tau^2}} \right).
\end{align*}
Comments.
You can make this question more general by assuming that
\begin{align*}
X=e^{\mu+\sigma Z_1} \quad \mbox{and} \quad Y = e^{\nu+\tau\big(\rho Z_1 + \sqrt{1-\rho^2} Z_2\big)},
\end{align*}
where $Z_1$ and $Z_2$ are two independent standard normal random variables, and $|\rho|<1$. In this case, the measure $Q$ can be defined by
\begin{align*}
\frac{dQ}{dP} = e^{-\frac{1}{2}\tau^2 + \tau \big(\rho Z_1 + \sqrt{1-\rho^2} Z_2\big)}.
\end{align*}
Moreover,
\begin{align*}
\widehat{Z}_1 = Z_1 - \rho\tau, \mbox{ and }
\widehat{Z}_2 = Z_2 - \sqrt{1-\rho^2}\tau
\end{align*}
are two standard independent normal random variables under the probability measure $Q$. See also this question at https://quant.stackexchange.com/questions/21361/how-to-use-a-change-of-numeraire-to-price-this-option/21375#21375.
Best Answer
The minimum of two independent exponential random variables with parameters $\lambda$ and $\eta$ is also exponential with parameter $\lambda+\eta$.
Also $\mathbb E\big[\min(X_1,X_2)+\max(X_1,X_2)\big]=\mathbb E\big[X_1+X_2\big]=\frac{1}{\lambda}+\frac{1}{\eta}$. Because $\mathbb E\big[\min(X_1,X_2)\big]=\frac{1}{\lambda+\eta}$, we get $\mathbb E\big[\max(X_1,X_2)\big]=\frac{1}{\lambda}+\frac{1}{\eta}-\frac{1}{\lambda+\eta}.$