Question

Let \(\mathrm{a}>0, \mathrm{~b}>0\), and \(\mathrm{f}\) a continuous strictly increasing function with \(\mathrm{f}(0)=0\). Prove that \(a b \leq \int_{0}^{a} f(x) d x+\int_{0}^{b} f^{-1}(x) d x\) Prove, moreover, that equality occurs if and on ly if \(\mathrm{b}=\mathrm{f}(\mathrm{a})\).

Short answer

Answer: The inequality holds for all continuous and strictly increasing functions \(f\) with \(f(0) = 0\). Equality occurs if and only if \(b = f(a)\).

Step-by-step solution

Recall properties of continuous strictly increasing functions

Since \(f\) is continuous and strictly increasing, its inverse function \(f^{-1}\) also exists and is continuous and strictly increasing. Moreover, as \(f(0) = 0\), we have \(f^{-1}(0) = 0\).

Apply the change of variables theorem to the second integral

Let \(u = f^{-1}(x)\). Then \(x = f(u)\) and \(dx = f'(u) du\). We can rewrite the second integral in terms of \(u\) as follows: $$\int_0^b f^{-1}(x) dx = \int_0^{f^{-1}(b)} u f'(u) du$$ We used the following substitutions: \(x=0 \Rightarrow u=f^{-1}(0)=0\) and \(x=b \Rightarrow u=f^{-1}(b)\). Note that since \(f^{-1}(x)\) is strictly increasing as well, \(\int_0^{f^{-1}(b)} u f'(u) du \geq 0\).

Rewrite the inequality in terms of one integral

Now, we can rewrite the inequality as: $$ab \leq \int_0^a f(x) dx + \int_0^{f^{-1}(b)} u f'(u) du$$ By rearranging the inequality, we get: $$\int_0^a f(x) dx - ab \geq -\int_0^{f^{-1}(b)} u f'(u) du$$

Prove the inequality using integration by parts

Let us apply integration by parts to the left-hand side of the inequality. Choose \(u=x\), \(dv=f(x) dx\), \(du=dx\), and \(v = \int_0^x f(t) dt\). Thus, we have: $$\int_0^a xf(x) dx - \int_0^a (\int_0^x f(t) dt) dx \geq -\int_0^{f^{-1}(b)} u f'(u) du$$ Observe that: $$\int_0^a (\int_0^x f(t) dt) dx = \int_0^a (\int_0^x f(x) dx)$$ Thus, the inequality becomes: $$\int_0^a xf(x) dx - a \int_0^a f(x) dx \geq -\int_0^{f^{-1}(b)} u f'(u) du$$ or equivalently, $$\int_0^a (x-a) f(x) dx \geq \int_0^{f^{-1}(b)} u f'(u) du$$ Now observe that the left-hand side of the inequality is non-negative. Since both sides involve non-negative quantities, we can apply the inequality on the integrals directly.

Prove when equality occurs

Equality occurs when both sides of the inequality are equal, i.e., when: $$\int_0^a (x-a) f(x) dx = \int_0^{f^{-1}(b)} u f'(u) du = 0$$ This is true if and only if the integrands are non-negative in each integral. Thus, we must have \((x - a)f(x) \geq 0\) for \(x\in[0, a]\) and \(0 \leq u \leq f^{-1}(b)\). This implies that \(x \leq a\) in the first condition, and since \(f(x)\) is strictly increasing, \(f(x) \leq f(a)\). So we have \(b = f(a)\). Conversely, if \(b=f(a)\), then both integrands are non-negative and the equality holds. In conclusion, the inequality is proven and the equality occurs if and only if \(b = f(a)\).