Auction Theory: Proving that the found equilibrium is indeed optimal

I have been looking at auction theory and in the book Auction Theory by Krishna, there is one (seemingly simple) inequality that I just cannot follow.

Context: given a private valuation $$x$$, the optimal bidding strategy has been found $$\beta(x)$$. Now, the author wants to show that behaving and bidding as if you were of type $$z$$, $$\beta(z)$$ does not increase profits. Then, calculating the difference between the profit in the optimum and the profit if you would behave as if you were of type $$z$$ leads to the following inequality. $$G(x)$$ being a probability distriubtion: $$\pi(\beta(x),x) - \pi( \beta(z),x) = G(z)(z-x) - \int_x^zG(y)dy \geq 0$$

The profit functions were calculated from a first-price auction in case it helps anyone. My question is why the inequality holds. Why is $$G(z)(z-x) - \int_x^zG(y)dy$$ larger than 0?

I hope you can help me :)

$$G(z) (z-x) = \int_x^z G(z) dy$$ and since $$G$$ is increasing on $$[x,z]$$, the right hand side is larger than $$\int_x^z G(y) dy$$.
• @Snowrabbit Same reason. In that case $G(z)$ is smaller than $G(y)$ for all $y \in [z,x]$ and $(z-x)<0$. Apr 22 at 6:28
By construction, $$\frac{\partial \pi}{\partial b}(b,x) = - G((\beta)^{-1}(b)) + (x-b) \frac{G'((\beta)^{-1}(b))}{(\beta)'((\beta)^{-1}(b))}\Bigg{|}_{b=\beta(x)}= 0,$$ where $$\frac{\partial \pi}{\partial b}(b,x)$$ is increasing in $$x$$.
Now consider some bid $$\widehat b<\beta(x)$$. By continuity of $$\beta$$, there is a type $$\widehat x such that $$\beta(\widehat x)=\widehat b$$. Hence, because $$\widehat x, $$\frac{\partial \Pi}{\partial b}(\widehat b,x) \geq \frac{\partial \Pi}{\partial b}(\widehat b, \widehat x) = \frac{\partial \Pi}{\partial b} (\beta(\widehat x),\widehat x) = 0.$$ Thus, the expected utility $$\Pi( b,x)$$ is increasing in $$b$$ for all $$b<\beta(x)$$. Analogously, $$\Pi(b,x)$$ is decreasing for all $$\widehat b'>\beta(x)$$.