Let $f:\mathbb{R}\to \mathbb{R}$ be a continuously differentiable function. Prove that for any $a.b\in \mathbb{R}$ $$\left (\int_a^b\sqrt{1+(f'(x))^2}\,dx\right)^2\ge (a-b)^2+(f(b)-f(a))^2$$.

source: This is from TIFR GS stage 2.

I think mean value theorem kills it but can't do it ...even try Cauchy-Schwarz inequality but nothing conclusion.Geometrically its true since distance between two point $(a,f(a))$ & $(b,f(b))$ is greater or equal to any arc length of same endpoints.

  • 3
    $\begingroup$ the smallest distance between the two points $(a, f(a))$ and $(b, f(b))$ is the straight line distance which is your RHS (the square root of that of course but same applies to LHS ; conclude... $\endgroup$
    – Conrad
    Mar 2, 2019 at 20:49
  • 1
    $\begingroup$ @Conrad But this is exactly what is to be proved, since the LHS is the definition of arc length. $\endgroup$ Mar 2, 2019 at 21:32
  • $\begingroup$ This is classic stuff - can do it locally using Taylor approximation so make the curve piecewise linear and use elementary geometry or as done in various answers with various inequalities $\endgroup$
    – Conrad
    Mar 3, 2019 at 0:22

4 Answers 4


Notice that the function $y \mapsto \sqrt{1+y^2}$ is strictly convex. So by the Jensen's inequality,

$$ \frac{1}{b-a} \int_{a}^{b} \sqrt{1 + f'(x)^2} \, \mathrm{d}x \geq \sqrt{1 + \left(\frac{1}{b-a}\int_{a}^{b} f'(x) \, \mathrm{d}x\right)^2} = \sqrt{1 + \left(\frac{f(b) - f(a)}{b-a} \right)^2}. $$

Multiplying both sides by $b-a$ and squaring proves the desired inequality. Moreover, by the strict convexity, the equality holds if and only if $f'$ is constant over $[a, b]$.

  • 1
    $\begingroup$ This really nice! $\endgroup$
    – Notsredt
    Mar 2, 2019 at 23:41

Note that for every complex valued integrable function $\phi :[a,b]\to \Bbb C$, it holds that $$ \left|\int_a^b \phi(x)\ dx\right|\le \int_a^b|\phi(x)|\ dx. $$ Let $\phi(x)=1+if'(x)$. Then we can see that $$\begin{align*} \left|\int_a^b \phi(x)\ dx\right|&=\left|(b-a)+i(f(b)-f(a))\right|\\&=\sqrt{(b-a)^2+(f(b)-f(a))^2} \end{align*}$$ and $$ \int_a^b|\phi(x)|\ dx=\int_a^b \sqrt{1+(f'(x))^2}\ dx. $$ Now, the desired inequality follows.
Note: The equality holds when $\text{arg}(\phi(x))$ is constant, that is, $\frac{f'(x)}{1}=f'(x)$ is constant.

  • 1
    $\begingroup$ (+1) Amazing, this should the accepted answer! Anyway, is there any reason to work with $\mathbb{C}$ rather than $\mathbb{R}^2$ with $\phi(x) = \gamma'(x)$ and $\gamma(x) = (x, f(x))$? $\endgroup$ Mar 3, 2019 at 0:12
  • $\begingroup$ Umm, sorry, I see no specific reason, since both are essentially the same version of the triangle inequality in integral form. But I just prefered $\Bbb C$-version because it can be easily derived from the real triangle inequality; if $f$ is real-valued, integrable, $\pm \int_a^b f\le \int_a^b |f|$. Thank you! $\endgroup$ Mar 3, 2019 at 0:21
  • 1
    $\begingroup$ This is slick. I wish I could upvote this answer twice.... $\endgroup$ Mar 3, 2019 at 1:48

An easy way to do this is to note that since distance is invariant under rotations, without loss of generality, we may assume that $f(a)=f(b).$ And now, since $\sqrt{1-f'(x)}\ge 0$ on $[a,b]$, the function in $C^1([a,b])$ that minimizes the integral coincides with the function $f$ that minimizes the integrand, and clearly, this happens when $f'(x)=0$ for all $x\in [a,b].$ That is, when $f$ is constant on $[a,b].$ Then, $f(x)=f(a)$ and the result follows.

If you want to do this without the wlog assumption, then argue as follows:

Let $\epsilon>0,\ f\in C^1([a,b])$ and choose a partition $P=\{a,x_1,\cdots,x_{n-2},b\}$.

The length of the polygonal path obtained by joining the points

$(x_i,f(x_i))$ is $\sum_i \sqrt{(\Delta x_i)^2+(\Delta y_i)^2}$ and this is clearly $\ge (b-a)^2+(f(b)-f(a))^2$. (You can make this precise by using an induction argument on $n$.)

And this is true for $\textit{any}$ partition $P$.

But the above sum is also $\sum_i\sqrt{1+\frac{\Delta y_i}{\Delta x_i}}\Delta x_i $ and now, upon applying the MVT, we see that what we have is a Riemann sum for $\sqrt{1+f'(x)}$.

To finish, choose $P$ such that $\left |\int^b_a\sqrt{1+f'(x)}dx- \sum_i\sqrt{1+f'(c_i)}\Delta x_i \right |<\epsilon $. (The $c_i$ are the numbers $x_i<c_i<x_{i-1}$ obtained from the MVT). Then,

$(b-a)^2+(f(b)-f(a))^2\le \sum_i\sqrt{1+f'(c)}\Delta x_i<\int^b_a\sqrt{1+f'(x)}+\epsilon.$

Since $\epsilon$ is arbitrary, the result follows.

For a slick way to do this, use a variational argument: assuming a minimum $f$ exists, consider $f+t\phi$ where $t$ is a real parameter and $\phi$ is arbitrary $C^1([a,b])$.

Subsitute it into the integral:

$l(t)=\int_a^b \sqrt{1+(f'+t\phi')^2}dx$.

Since $f$ minimizes this integral, the derivative of $l$ at $t=0$ must be equal to zero. Then,

$0=l'(0)= \int_a^b \dfrac{f'\phi'}{\sqrt{1+(f')^2}}dx$.

After an integration by parts, we get

$\dfrac{f'}{\sqrt{1+(f')^2}} = c$ for some constant $c\in \mathbb R,$ from which it follows that $f'=c$. And this means, of course, that the graph of $f$ is a straight line connecting $(a,f(a))$ and $(b,f(b)).$ The desired inequality follows.


Expanding upon what @Conrad said, the shortest distance between two points is the distance of the line between, which is what your RHS is measuring (it is actually the square of the distance from $(a, f(a))$ to $(b, f(b))$.

Now if we assume $\left (\int_a^b\sqrt{1+(f'(x))^2}\,dx\right)^2 < (a-b)^2+(f(b)-f(a))^2$, then we have contradicted the fact that that the shortest distance between $(a, f(a))$ and $(b, f(b))$ is $\sqrt{(a-b)^2+(f(b)-f(a))^2}$. Therefore, it must be the case that $\left (\int_a^b\sqrt{1+(f'(x))^2}\,dx\right)^2 \geq (a-b)^2+(f(b)-f(a))^2$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.