I'd like to prove the following:

$$ a + b + 4 \sqrt{1 + a^{2} + b^{2}} \leq 4 \sqrt{a^{2} + b^{2}} + \sqrt{1+b^{2}} + \sqrt{1+a^{2}} + 2 $$

for all $a, b \in \mathbb{R}_{>0}$.

Question: is there a way to prove this inequality?

Motivation: In a paper, Borm et al. work out the theory behind stochastic cooperative game theory, which is a generalisation of the classical (deterministic) cooperative game theory developed in the 1950s. For my master's thesis, I am currently trying to apply their model to production prediction problems for renewable energy sources. Owners of renewable energy sources need to report how much energy their wind mills or solar panels they think will produce the next hour/day to the network operator. Often, they need to pay a fine when the actual production levels deviate from their productions.

In my thesis, I assume that the normalized error distribution of the predictions is a normal distribution $N(0, \sigma^{2})$. Furthermore, I assume that the costs of the fine increase linearly with the absolute value of the error made (as is the case in Spain: see this article). So the fine/cost distribution is a so-called half-normal distribution $H(0, \sigma^{2})$.

The owners of the renewable energy sources can cooperate to reduce the fine they need to pay. In one variant of cooperation, they can combine their predictions to minimise their total fine, after which they distribute the costs among all owners. In the three-player game, I assume that owner 1 has an error distribution that goes like $X_{1} \sim N(0, \sigma_{1}^{2})$. For owner 2 we have $X_{2} \sim N(0, \sigma_{2}^{2})$ and for owner 3 we have $X_{3} \sim N(0, \sigma_{3}^{2})$. Furthermore, I assume that $\sigma_{2} = a \sigma_{1} = a \sigma$, and $\sigma_{3} = b \sigma_{1} = b \sigma$. So for our cost game, the stochastic costs are distributed as follows:

$$R(S) = \left\{ \begin{array}{lllllll} |X_{1}| & \mbox{for } S = \{1\} \\ |X_{2}| & \mbox{for } S = \{2\} \\ |X_{3}| & \mbox{for } S = \{3\} \\ |X_{1} + X_{2}| & \mbox{for } S= \{1,2\} \\ |X_{1} + X_{3}| & \mbox{for } S = \{1,3\} \\ |X_{2} + X_{3}| & \mbox{for } S = \{2,3\} \\ |X_{1} + X_{2} + X_{3}| & \mbox{for } S= \{1,2,3\} = N \end{array} \right.$$

Keep in mind that for normally distributed random variables $X_{1}$ and $X_{2}$, we can compute the convolution: $$X_{1} + X_{2} \sim N(0, \sigma_{1}^{2} + \sigma_{2}^{2}) = N(0, \sigma^{2}(1+a^{2})).$$ This means $$|X_{1} + X_{2}| \sim H(0, \sigma^{2}(1+a^{2})).$$ Analogously, we can compute the convolutions for the other coalitions, too.

Now, I assume that that all players of the game have expectational preferences (see page 5 of the paper by Borm et al.). This means that for a player $i$, we have $X \succsim_{i} Y$ if and only if $E(X) \geq E(Y)$.

Players compare the stochastic payoffs with their preferences and the so-called tracking function (my terminology, not theirs). If player $i$ receives $p_{i}$th part of the payoff (or costs, in my case) of the stochastic coalitional value $R(S)$, there exists a number $\alpha_{i}$ such that $p_{i} R(S) \sim_{i} \alpha_{i} R(T)$ for the set $T \supset S$. Notice that $\alpha_{i}$ depends on $S$, $T$ and $p_{i}$, so it is a function of these entities. This is the tracking function. In case all players have expectational preferences, we have $$\alpha_{i} (S, T, p) = p \frac{E(R(S))}{E(R(T))}$$ for $i=1,2,3$.

The core of a stochastic cooperative game can be computed too. It consists of those efficient (which means that $\sum_{i} p_{i} = 1$) allocations $p$ such that $\sum_{i \in S} p_{i} / \alpha_{i} (S, N, 1) \geq 1$ for all coalitions $S \in 2^{N}$. For a cost game (my situation) the inequalities are flipped. I computed the core for the game I described above. It consists of those efficient allocations $p \in \mathbb{R}^{3}$ such that: $$p_{1} \leq \frac{1}{\sqrt{1+a^{2}+b^{2}}}, \quad p_{2} \leq \frac{a}{\sqrt{1+a^{2}+b^{2}}}, \quad p_{3} \leq \frac{b}{\sqrt{1+a^{2}+b^{2}}},\\ p_{1} + p_{2} \leq \sqrt{\frac{1+a^{2}}{1+a^{2}+b^{2}}}, \quad p_{1} + p_{3} \leq \sqrt{\frac{1+b^{2}}{1+a^{2}+b^{2}}}, \quad p_{2} + p_{3} \leq \sqrt{\frac{a^{2}+b^{2}}{1+a^{2}+b^{2}}}. $$

I then calculated the Shapley value for this game. The procedure for the computation of this value is explained on pages 9 and 10 of Borm et al.'s paper. It is as follows: \begin{align*} \phi (\alpha) &= \frac{1}{n!} \sum_{\sigma \in \Pi(N)} m^{\sigma} (\alpha) \\ &= \frac{1}{3!} \big{(} m^{(1,2,3)} + m^{(1,3,2)} + m^{(2,1,3)} + m^{(2,3,1)} + m^{(3,1,2)} + m^{(3,2,1)} \big{)} \\ &= \frac{1}{6\sqrt{1+a^{2}+b^{2}}} \Big{(} 2 + \sqrt{1+a^{2}} + \sqrt{1+b^{2}} + 2 \sqrt{1+a^{2}+b^{2}} - 2 \sqrt{a^{2}+b^{2}} - a - b, \\ & \qquad \qquad \qquad \quad 2a + \sqrt{a^{2} + 1} + \sqrt{a^{2}+b^{2}} + 2 \sqrt{1+a^{2}+b^{2}} - 2 \sqrt{1+b^{2}} - 1 - b, \\ & \qquad \qquad \qquad \quad 2b + \sqrt{b^{2}+1} + \sqrt{b^{2}+a^{2}} + 2 \sqrt{1+a^{2}+b^{2}} - 2 \sqrt{1+a^{2}} - 1 - a \Big{)} \end{align*}

I am now trying to verify whether this Shapley value lies in the core of the game. In particular, I am trying to verify the sixth (and last) inequality of the core. So if we take the second and third coordinates of the Shapley value and add them together, then this sum mustn't exceed the value $\sqrt{\dfrac{a^{2}+b^{2}}{1+a^{2}+b^{2}}} $. With a little bit of basic algebra, you will arrive at the inequality at the top of this question.

What I tried: I already tried applying the triangle inequality and the AM-GM inequality, but that did not seem to work. I also asked a similar question on MSE on the fourth inequality for the determination of whether the Shapley value is in the core. Two users on that page were kind enough to provide me with an answer to that question. However, I don't think I can generalise Alex Francisco's answer to this inequality, and I don't understand Piquito's answer fully yet. I wonder whether there are other ways to approach this problem.


This rewrites as $4f(a^2 +b^2)\le f(a^2) +f(b^2 )+2f(0) $ for a decreasing function $f(x)=\sqrt{x+1}-\sqrt{x}=1/(\sqrt{x+1}+\sqrt{x})$.

  • $\begingroup$ That was quick. Thank you for your succinct and elementary proof! $\endgroup$
    – Max Muller
    Mar 13 '18 at 19:15

Because $$2+\sqrt{1+a^2}+\sqrt{1+b^2}-a-b=2+\frac{1}{a+\sqrt{1+a^2}}+\frac{1}{b+\sqrt{1+b^2}}\geq$$ $$\geq\tfrac{2}{\sqrt{a^2+b^2}+\sqrt{1+a^2+b^2}}+\tfrac{1}{\sqrt{a^2+b^2}+\sqrt{1+a^2+b^2}}+\tfrac{1}{\sqrt{a^2+b^2}+\sqrt{1+a^2+b^2}}=$$ $$=4\sqrt{1+a^2+b^2}-4\sqrt{a^2+b^2}.$$


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