# Geometric intuition for the inequality $(f(y) - c) ( y - d ) \geq (f(d) - c) ( f^{-1}(c) - d )$

Good day to everyone. I am interested in the geometric intuition for the following statement:

Let $f:\mathbb{R} \mapsto \mathbb{R}$ be a monotonically increasing, invertible function and $c,d \in \mathbb{R}$. Then for any $y \in \mathbb{R}$ we have $$\Big(f(y) - c\Big) \Big( y - d \Big) \geq \Big(f(d) - c\Big) \Big( f^{-1}(c) - d \Big).$$

It is a lemma which I have encountered in the article (Lemma 3.1):
T. Kerkhoven and J. W. Jerome. $L_\infty$ stability of finite element approximations of elliptic gradient equations. Numerische Mathematik, 57:561, 1990.

The lemma is proven in the paper. The proof is analytic and it is rather elementary. However, it seems to me that this statement is releated to some geometric property of graphs of real increasing functions. And I am interested in this geometric intuition. Perhaps it is a standard argument, but I cannot see it. Thus my question is:

What are the geometrical reasons for the statement cited above?

Some thoughts:
It is maybe relevant to multiply the inequality by (-1), reversing its direction. It is so, because if we take $f(x):=x$, then also $f^{-1}(x) = x$ and we obtain $LHS:=(y-c)(y-d) \geq -(d-c)^2=:RHS$, so RHS is negative for all $c \neq d$. Thus if LHS is positive, then the signs differ and probably no geometrical argument can be given. So the interesting case would be when LHS is also negative.

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I get very confused thinking geometrically with $y$ on the horizontal axis, so let us call the variable $x$ instead, and let $y = f(x)$. Also, let's say $x_1 = f^{-1}(c)$, $y_1 = c$, and $x_2 = d$, $y_2 = f(d)$, so that the points $(x_1,y_1)$ and $(x_2,y_2)$ lie on the monotonically increasing curve $y=f(x)$. The inequality becomes $$(y-y_1)(x-x_2)\ge(y_2-y_1)(x_1-x_2).$$ As $f$ is increasing, $y_2-y_1$ has the same sign as $x_2-x_1$, so the right-hand side is always negative. If $x$ is outside the interval between $x_1$ and $x_2$, the left-hand side is positive, so the inequality holds. The interesting part is when $x$ lies between $x_1$ and $x_2$, and here we will look at it geometrically.

Negate the inequality to obtain $$(y-y_1)(x_2-x)\le(y_2-y_1)(x_2-x_1).$$ Assuming $x_1<x_2$, the right-hand side is the area of the rectangle $[x_1,x_2]\times[y_1,y_2]$, while the left-hand side is the area of $[x,x_2]\times[y_1,y]$. The latter rectangle is a subset of the former, so the inequality holds. If $x_1>x_2$, negate all four terms and you again get the positive areas of two rectangles, and the same reasoning applies.

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Thank you. Your argument is really nice. Therefore RHS is the area of some fixed rectangle $S:=[x_1,x_2] \times [y_1,y_2]$, and LHS is the area of rectangle $S_x:=[x,x_2]\times[y_1,y]$. And depending on $x$ it may be that $S_x \subset S$, so its area is smaller, or $S_x \not \subset S$, but then its area is negated, so the inequality holds for the both cases. – Konrad Sakowski Aug 3 '12 at 8:14
Yes, I guess I didn't need to discard cases before moving to the geometrical argument. Thanks for your nice observation. – Rahul Aug 3 '12 at 8:29

Put $f^{-1}(c)=:u$. Then we have to prove that $$\bigl(f(y)-f(u)\bigr)(y-d)\geq\bigl(f(d)-f(u)\bigr)(u-d)$$ or $$\bigl(f(y)-f(u)\bigr)(d-y)\leq\bigl(f(d)-f(u)\bigr)(d-u)\ .$$ Here the RHS is positive. When $y<\min\{u,d\}$ or $y>\max\{u,d\}$ then the LHS is negative. When $y$ is between $u$ and $d$ then the LHS is also positive, but both factors are less in absolute value than the coresponding factors on the RHS.

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Thank you. It is a very elegant proof. However, I was mostly concerned about some geometric argument, possibly releated to the graph of the function $f$. – Konrad Sakowski Aug 3 '12 at 9:10
@KonradSakowski: Perhaps this diagram might help provide some geometric insight. In the case where the product on the left is positive, the product on the left is the area of the red rectangle, whereas the product on the right is the area of the green rectangle. – robjohn Aug 3 '12 at 13:37
Christian: feel free to use the diagram linked in my previous comment. – robjohn Aug 3 '12 at 13:39