Does is it change an auction's incentives when causing the winner to pay more makes losers pay less? Say there are three roommates moving into an apartment with three rooms. Two of the apartment's rooms are identical, but the third one is valued higher by all three parties (say it's bigger and has a private bathroom). To decide which one gets the big room, the roommates decide to hold an auction.
If the apartment's monthly rent is $p$ dollars in total, the idea behind the auction is to find a number $\delta$ such that the winner of the auction pays $\frac{p}{3} + \delta$, while the two losers pay $\frac{p}{3} - \frac{\delta}{2}$. In other words, the winner of the auction (who gets the big room) pays a bigger share of the rent compared to the two losers.
Does this type of auction—where the two losers benefit from the value of $\delta$ being large—differ from traditional auctions, where it doesn't concern the losers what the winner pays? For example, does it still apply that if the price is decided using an English auction, that each bidder $i$ is incentivized to never bid higher than their value $v_i$? Intuitively, you might think that if the bidder $s$ with the second highest value $v_s$ for $\delta$ is reasonably confident that the to-be winner $w$ has a much higher value $v_w$ for $\delta$ they might bid higher than their own value $v_s$, in hopes of raising $\delta$ and lowering the rent that they have to pay.
 A: In your auction, losing is not as bad as it is in an English auction.  Therefore you should expect that in your auction, people bid lower, not higher.
To check this, I'll take the three reservation prices to be drawn independently and uniformly from $[0,1]$.  (For other distributions, it will be easy to modify these calculations.)
1)  If your reservation price is $x$ and your bid is $b$, your
expected return is 
$$\pi(x,b)=Prob(Win)(x-b)+E(b)$$
where $E(b)$ is your expected gain if you lose the auction.  ($E$ can't depend on $x$ because $x$ is not observable to anyone but you.)
2)  Let $b=B(x)$ be a symmetric monotonic Nash equilibrium bidding strategy. (Symmetric means that all three players employ the same strategy.)
Let $\pi_0(x)=\pi(x,B(x))$.
3)  By the chain rule,
$$\pi_0'(x)={\partial\pi\over\partial x}+{\partial\pi\over\partial b}B'(x)$$
But when you choose your bid optimally, the second partial is zero, so we have 
$$\pi_0'(x)={\partial\pi\over\partial x}=Prob(Win)=x^2$$
with the last equality following from the  uniformity assumption.
Clearly $\pi_0(0)=1/3$, so we have
$$\pi_0(x)={x^3\over 3}+{1\over 3}$$
4)  Directly from the definitions, and continuing to assume we're in a symmetric monotonic Nash equilibrium, we have
$$\pi_0(x)=\pi(x,B(x))=x^2(x-B(x))+E(B(x))$$
Combining this with (3), we get
$${x^3\over 3}+{1\over 3}=x^2(x-B(x))+E(B(x))\hskip{2pc}(1)$$
where
$$ 2E(B(x))=\int_0^1\int_0^1 max(B(y),B(z))dydz- \int_0^x
\int_0^x max(B(y),B(z)))dz$$
Equation (1) is an integral equation for $B$.  Solving it, I get $B(x)=x/2$.   In particular, your optimal bid is always less than your reservation price.
