Laisant’s Identity at Work

October 24, 2026October 24, 2026 / Michael Xue / Leave a comment

In my earlier post, “A Gem from Charles-Ange Laisant”, I introduced an identity involving a function and its inverse.

Let’s return to the identity marked (*) in “Meeting Young’s Theorem”:

$\displaystyle \int_0^a f(x)\; dx+\int_0^{f(a)}g(y)\;dy=af(a)\quad\quad\quad(1)$

where $f$ and $g=f^{-1}$ satisfy the assumptions stated in Young’s theorem.

How can we obtain this identity using Laisant’s result?

Recall that Laisant’s identity gives

$\displaystyle \int_\alpha^\beta f(x)\; dx = \beta f(\beta)-\alpha f(\alpha)-[F(f(\beta)) - F(f(\alpha))]$

where $\quad F' = f^{-1}$ .

Let

$\displaystyle F(y) = \int_0^y f(x)\;dx$

so that

$F'(y) = g^{-1}(y)=f(y)$ .

Applying Laisant’s identity to $g$ , we obtain

$\displaystyle \int_{0}^{f(a)} g(y)\;dy = f(a)\cdot g(f(a))-0\cdot g(0) - [F(g(f(a)))-F(g(0))]$

$\displaystyle = f(a)\cdot a - \int_0^{g(f(a))}f(x)\;dx + \int_0^{g(0)} f(x)\;dx$ .

Since $g(f(a)) = a$ and $g(0)=0$ , this becomes

$\displaystyle \int_{0}^{f(a)} g(y)\;dy=af(a) - \int_0^a f(x)\; dx.$

Therefore,

$\displaystyle \int_0^a f(x) \;dx + \int_0^{f(a)} g(y)\;dy = af(a)$ .

There it is: the identity we needed in our proof of Young’s theorem, obtained directly from Laisant’s result.

From Hölder to Minkowski

October 21, 2026October 24, 2026 / Michael Xue / Leave a comment

Let’s use Hölder’s inequality to prove another inequality, one that looks deceptively familiar:

$\displaystyle \left(\int_a^b |f(x)+g(x)|^p dx\right)^{1/p} \leq \left(\int_a^b |f(x)|^p dx\right)^{1/p}+\left(\int_a^b |g(x)|^p dx\right)^{1/p}.$

This is Minkowski’s inequality. It holds for continuous functions $f, g$ on $[a,b]$ and $p \ge 1.$

Looking at the inequality, you might recognize something like the triangle inequality. The absolute value of a sum is no greater than the sum of the absolute values:

$\displaystyle |f+g|\leq |f|+|g|.\quad\quad\quad(1)$

But how do we get from this simple observation to Minkowski’s inequality?

Let’s see.

The case $p=1$ is immediate. Integrating the ordinary triangle inequality gives

$\begin{aligned} \int_a^b |f(x)+g(x)| \;dx&\leq \int_a^b \bigl(|f(x)|+|g(x)|\bigr) \;dx = \int_a^b |f(x)|\;dx+\int_a^b |g(x)|\;dx\end{aligned}.$

That is Minkowski’s inequality for $p=1.$

The interesting case is $p>1$ . Here we will call on Hölder’s inequality.

Let $q>1$ satisfy

$\frac1p+\frac1q=1.$

Equivalently,

$1-\frac{1}{q} = \frac{1}{p}\quad\quad\quad(2)$

so that

$(p-1)q = p.\quad\quad\quad(3)$

Multiplying (1) by $|f+g|^{p-1},$

$|f +g||f+g|^{p-1} \le |f||f+g|^{p-1} + |g||f+g|^{p-1}$

gives

$|f+g|^p \le |f||f+g|^{p-1} + |g||f+g|^{p-1}.$

Integrating both sides,

$\displaystyle \int_a^b |f+g|^p\;dx \le \underline{\int_a^b |f||f+g|^{p-1}\;dx} + \underline{\int_a^b |g||f+g|^{p-1}\;dx}.$

By Hölder’s inequality,

$\displaystyle \underline{\int_a^b |f||f+g|^{p-1}\;dx} \le \left(\int_a^b |f|^p\;dx\right)^\frac{1}{p}\cdot \left(\int_a^b|f+g|^{(p-1)q}\;dx \right)^\frac{1}{q},$

$\displaystyle \underline{\int_a^b |g||f+g|^{p-1}\;dx} \le \left(\int_a^b |g|^p\;dx \right)^\frac{1}{p}\cdot \left(\int_a^b|f+g|^{(p-1)q}\;dx \right)^\frac{1}{q}.$

Hence,

$\displaystyle \int_a^b |f+g|^p\;dx \le \left(\int_a^b |f+g|^{(p-1)q}\;dx\right)^\frac{1}{q}\left(\left(\int_a^b |f|^p\;dx \right)^\frac{1}{p}+\left(\int_a^b|g|^p\;dx\right)^\frac{1}{p}\right).\quad\quad\quad(4)$

Denoting

$\displaystyle I=\int_a^b |f+g|^p\;dx,$

(4) becomes

$\displaystyle I \overset{(3): (p-1)q=p}{\le} I^\frac{1}{q}\cdot\left(\left(\int_a^b |f|^p\;dx\right)^\frac{1}{p}+\left(\int_a^b|g|^p\;dx\right)^\frac{1}{p}\right).$

If $I=0$ , the result is immediate. Otherwise, divide both sides by $I^\frac{1}{q}$ to obtain

$I^{1-\frac{1}{q}} \le \left(\left(\int_a^b |f|^p\;dx\right)^\frac{1}{p}+\left(\int_a^b|g|^p\;dx\right)^\frac{1}{p}\right)\overset{(2): 1-\frac{1}{q}=\frac{1}{p}}{\implies} I^\frac{1}{p} \le \left(\left(\int_a^b |f|^p\;dx\right)^\frac{1}{p}+\left(\int_a^b|g|^p\;dx\right)^\frac{1}{p}\right).$

Therefore,

$\displaystyle \left(\int_a^b |f(x)+g(x)|^p dx\right)^{1/p} \leq \left(\int_a^b |f(x)|^p dx\right)^{1/p}+\left(\int_a^b |g(x)|^p dx\right)^{1/p}.$

We began with the ordinary triangle inequality. It took us part of the way, but not all the way. Hölder supplied what was missing.

One inequality leads to another.

And now we have a new question: what can Minkowski’s inequality do for us?

Where Young’s Inequality Shines

October 19, 2026October 21, 2026 / Michael Xue / Leave a comment

We have proved the Cauchy–Schwarz inequality in two ways. One proof uses Young’s inequality; another starts with the elementary fact $(a-b)^2\geq 0$ .

Can we generalize Cauchy–Schwarz to other exponents?

Yes. This leads to Hölder’s inequality.

For $p,q>1$ satisfying

$\displaystyle \frac1p+\frac1q=1$ ,

Hölder’s inequality states that

$\displaystyle \int_a^b |fg|\; dx \leq \left(\int_a^b |f|^p\; dx\right)^{1/p} \left(\int_a^b |g|^q \;dx\right)^{1/q}$

for continuous functions $f,g$ on $[a,b]$ .

Taking $p=q=2$ recovers the Cauchy–Schwarz inequality.

Recall Young’s inequality:

$\displaystyle uv\leq \frac{u^p}{p}+\frac{v^q}{q}, \qquad u,v\geq 0$ .

Let

$\displaystyle A=\left(\int_a^b |f|^p\; dx\right)^{1/p},\qquad B=\left(\int_a^b |g|^q\; dx\right)^{1/q}$ .

Assuming $A,B>0$ , define

$\displaystyle u(x)=\frac{|f(x)|}{A}, \qquad v(x)=\frac{|g(x)|}{B}$ .

Then

$\displaystyle \int_a^b u^p\; dx=1,\qquad \int_a^b v^q\; dx=1$ .

Applying Young’s inequality pointwise and integrating, we obtain

$\displaystyle \begin{aligned} \int_a^b uv\; dx &\leq \frac1p\int_a^b u^p\; dx+\frac1q\int_a^b v^q\; dx \\ &=\frac1p+\frac1q \\ &=1.\end{aligned}$

Since

$\displaystyle uv=\frac{|fg|}{AB}$ ,

it follows that

$\displaystyle \int_a^b |fg|\; dx\leq AB$ .

Substituting the definitions of $A$ and $B$ proves Hölder’s inequality. If either $A$ or $B$ is zero, the result is immediate.

We have come a long way from Cauchy–Schwarz. The elementary inequality $(a-b)^2\geq 0$ gives us a simple proof of that result. But Young’s inequality takes us further, providing a direct proof of the more general Hölder’s inequality.

Excuse Me, Professor, Where Did That Rabbit Come From?

October 19, 2026October 19, 2026 / Michael Xue / Leave a comment

A long time ago, I learned the Cauchy–Schwarz inequality from the quadratic proof. It goes like this:

For continuous functions $f$ and $g$ on $[a,b]$ , start with

$\displaystyle \int_a^b (f-\lambda g)^2 \; dx\geq 0$ .

Expand. We get a quadratic in $\lambda$ . Its discriminant cannot be positive. Rearrange, and voilà!

$\displaystyle \left(\int_a^b fg \; dx\right)^2 \leq \left(\int_a^b f^2 \; dx\right) \left(\int_a^b g^2 \; dx\right)$ .

Q.E.D.

Everyone nodded, everyone cheered.
I looked around: What just happened here?

Don’t get me wrong. The proof is perfect and beautiful. But I can’t help wondering how someone came up with $(f-\lambda g)^2$ in the first place.

Was there a flash of inspiration? A lucky guess? Seventeen failed attempts and a very large wastebasket?

The proof doesn’t say.

It simply produces a rabbit from a hat and asks us to admire the rabbit.

I do admire it. I just want to know how it got in there.

A proof can show me that the trick works. Can it also show me why I should have thought of the trick?

All these years later, I am still looking for the rabbit’s entrance.

Cauchy-Schwarz: A Different Proof

October 19, 2026October 19, 2026 / Michael Xue / Leave a comment

In “Cauchy-Schwarz: A Young Proof”, we set out to prove the Cauchy-Schwarz integral inequality using Young’s inequality with $p=q=2$ .

To make it work, we defined our normalized functions using absolute values.

Why? Because Young’s inequality requires nonnegative inputs.

But that raises a question: Did we really need those absolute values?

Consider a simple algebraic fact:

For any real numbers $A$ and $B$ , $(A-B)^2 \ge 0$ .

Expanding and rearranging gives

$\displaystyle AB \le \frac{A^2+B^2}{2}$ .

Notice that $A$ and $B$ can be positive, negative, or zero. No absolute values are needed.

Let’s see how this leads to a different proof.

We want to prove the Cauchy-Schwarz integral inequality:

$\displaystyle \left|\int_a^b f(x)g(x)\; dx\right| \le \sqrt{\int_a^b f^2(x)\;dx}\sqrt{\int_a^b g^2(x)\; dx}$ .

Assume $f$ and $g$ are continuous on $[a,b]$ .

If either

$\displaystyle \int_a^b f^2(x)\; dx=0$

$\displaystyle \int_a^b g^2(x)\; dx=0$ ,

then the corresponding function is identically zero on $[a,b]$ . Consequently, $\int_a^b f(x)g(x)\;dx=0$ , and the desired inequality holds.

We may therefore assume both integrals are strictly positive.

Since both integrals are strictly positive, we can define

$\displaystyle F(x)=\frac{f(x)}{\sqrt{\int_a^b f^2(x)\;dx}},\qquad G(x)=\frac{g(x)}{\sqrt{\int_a^b g^2(x)\; dx}}$ .

By design,

$\displaystyle \int_a^b F^2(x)\;dx=1,\qquad \int_a^b G^2(x)\;dx=1$ .

Notice that we did not need absolute values. The normalized functions $F$ and $G$ can take either positive or negative values.

For every $x\in [a,b]$ , the inequality

$\displaystyle F(x)G(x) \leq \frac{F^2(x)+G^2(x)}{2}$

holds because

$\displaystyle (F(x)-G(x))^2\geq 0$ .

Integrating both sides over $[a,b]$ , we obtain

$\displaystyle \int_a^b F(x)G(x)\; dx \leq \frac12\int_a^b \bigl(F^2(x)+G^2(x)\bigr)\;dx$ .

Using our normalization,

$\displaystyle \begin{aligned} \int_a^b F(x)G(x)\;dx & \leq \frac12(1+1) \\ &=1.\end{aligned}$

Substituting the definitions of $F$ and $G$ gives

$\displaystyle \frac{\int_a^b f(x)g(x)\; dx}{\sqrt{\int_a^b f^2(x)\;dx}\sqrt{\int_a^b g^2(x) dx}} \le 1$ .

Multiplying both sides by the positive denominator, we arrive at

$\displaystyle \int_a^b f(x)g(x)\; dx \leq \sqrt{\int_a^b f^2(x)\;dx}\sqrt{\int_a^b g^2(x)\; dx}$ .

The inequality above gives an upper bound. Applying the same argument to $-f$ and $g$ gives the corresponding lower bound. We get

$\displaystyle -\int_a^b f(x)g(x) \; dx \le \sqrt{\int_a^b f^2(x)\; dx}\sqrt{\int_a^b g^2(x)\;dx}$ .

Combining the two inequalities gives

$\displaystyle \left|\int_a^b f(x)g(x)\; dx\right| \leq \sqrt{\int_a^b f^2(x) \; dx}\sqrt{\int_a^b g^2(x)\; dx}$ .

This completes the proof of the Cauchy-Schwarz integral inequality.

Mathematics often rewards us for looking at a familiar result from a different angle.

Sometimes, all it takes is a simple question to find a different way forward.

The destination is the same. The proof is different. And that difference is worth seeing.

Exercise 1. Prove that $(\forall x, y \in \mathbb{R}, |x| \le y) \iff -y \le x \le y$ . This is not a joke!

Cauchy-Schwarz: A Young Proof

October 16, 2026October 16, 2026 / Michael Xue / 1 Comment

In “Problem of the Week: See Cauchy-Schwarz in Action“, we turned to the Cauchy-Schwarz inequality:

$\displaystyle \bigg|\int_a^b fg\;dx \bigg| \le \sqrt{\int_a^b f^2\;dx}\sqrt{\int_a^b g^2\;dx}$

where $f, g$ are continuous functions on $[a, b]$ .

Let’s prove it using Young’s inequality.

If either $\int_a^b f^2\;dx = 0$ or $\int_a^b g^2\;dx =0$ , then the desired inequality holds trivially. We may therefore assume both integrals are positive.

Define

$\displaystyle F = \frac{|f|}{\sqrt{\int_a^b f^2\;dx}}, \quad G = \frac{|g|}{\sqrt{\int_a^b g^2\;dx}}$ .

Since $F \ge 0$ and $G \ge 0$ for all $x \in [a,b]$ , we can apply Young’s inequality with $p=q=2$ :

$\displaystyle \frac{F^2}{2} + \frac{G^2}{2} \ge FG$ .

Integrating both sides, we obtain

$\displaystyle \int_a^b \left(\frac{F^2}{2} + \frac{G^2}{2}\right) \; dx \ge \int_a^b FG\;dx$ .

That is,

$\displaystyle \frac{1}{2}\left(\int_a^b F^2\;dx + \int_a^b G^2\;dx\right) \ge \int_a^b FG\;dx.\quad\quad\quad(*)$

By the definitions of $F$ and $G$ ,

$\displaystyle \begin{aligned} \int_a^b F^2\;dx &= \int_a^b \left(\frac{|f|}{\sqrt{\int_a^b f^2\;dx}}\right)^2\;dx \\ &= \int_a^b \frac{|f|^2}{\int_a^b f^2\;dx}\;dx \\ &=\frac{\int_a^b f^2\;dx}{\int_a^b f^2\;dx} \\ &= 1,\end{aligned}$

$\displaystyle \begin{aligned} \int_a^b G^2\;dx &= \int_a^b \left(\frac{|g|}{\sqrt{\int_a^b g^2\;dx}}\right)^2\;dx \\ &= \int_a^b \frac{|g|^2}{\int_a^b g^2\;dx}\;dx \\ &=\frac{\int_a^b g^2\;dx}{\int_a^b g^2\;dx} \\ &= 1.\end{aligned}$

Substituting into (*) yields

$\displaystyle \frac{1}{2}(1 +1) \ge \int_a^b \frac{|f||g|}{\sqrt{\int_a^b f^2\;dx}\sqrt{\int_a^b g^2\;dx}} \; dx$

which gives

$\displaystyle 1 \ge \frac{1}{\sqrt{\int_a^b f^2\;dx}\sqrt{\int_a^b g^2\;dx}}\int_a^b \underbrace{|f||g|}_{|fg|}\;dx$ .

Therefore,

$\displaystyle \int_a^b |fg|\;dx \le \sqrt{\int_a^b f^2\;dx}\sqrt{\int_a^b g^2\;dx}$ .

Finally, since

$\displaystyle \int_a^b |fg| \;dx \ge \bigg|\int_a^b fg\;dx\bigg|$ ,

we conclude that

$\displaystyle \bigg|\int_a^b fg \;dx \bigg| \le \sqrt{\int_a^b f^2\;dx}\sqrt{\int_a^b g^2\;dx}$ .

This is the Cauchy-Schwarz inequality.

Exercise 1. Explain why the desired inequality holds trivially if either

$\displaystyle \int_a^b f^2\;dx =0$ or $\displaystyle \int_a^b g^2\;dx =0$ .

Exercise 2. Prove that $|x||y| = |xy|$ for all $x, y \in \mathbb{R}$ . This is not a joke!

Problem of the Week: See Cauchy-Schwarz in Action

October 14, 2026October 14, 2026 / Michael Xue / 1 Comment

Assume $f$ is continuous on $[a, b]$ . Prove:

$\displaystyle \int_a^b f(x) \;dx \le \sqrt{b-a}\sqrt{\int_a^b f^2(x)\;dx}$ .

Since

$\displaystyle \int_a^b f(x)\;dx = \int_a^b f(x)\cdot 1\;dx$ ,

by Cauchy-Schwarz inequality,

$\displaystyle \bigg|\int_a^b f(x)\cdot 1\;dx \bigg| \le \sqrt{\int_a^b 1^2\;dx} \sqrt{\int_a^b f^2(x)\;dx} = \sqrt{b-a}\sqrt{\int_a^b f^2(x)\;dx}$ .

It follows that

$\displaystyle \int_a^b f(x)\; dx \le \bigg|\int_a^b f(x) \;dx \bigg| \le \sqrt{b-a}\sqrt{\int_a^b f^2(x)\;dx}$ .

Exercise 1. Show that $\forall x \in \mathbb{R}, x \le |x|$ . This is not a joke!

Meeting Young’s Theorem

October 12, 2026October 12, 2026 / Michael Xue / 1 Comment

In my first post, “An Inequality Hiding in an Integral”, we discovered Young’s inequality hiding inside an integral inequality.

We also asked:

Where does this integral inequality come from?

In the second post, “Young’s Inequality: Just Find the Minimum”, we proved Young’s inequality.

But that still left our original question unanswered.

It has been lingering ever since.

So, in this post, we finally answer it.

And the answer takes us to:

Young’s theorem

Let $f:[0, \infty) \rightarrow [0, \infty)$ be continuous on $[0, \infty)$ , continuously differentiable on $(0, \infty)$ , strictly increasing and onto, with

$f(0) = 0$ ,

and let

$g = f^{-1}$ .

For $a, b \ge 0$ , Young’s theorem says

$\boxed{ab \le \int_{0}^{a} f(x) \;dx + \int_{0}^{b} g(y)\;dy}$ .

Equality occurs when

$b = f(a)$ ,

or equivalently,

$a = g(b)$ .

This theorem is more general than the integral inequality we started with.

Let’s see it.

Draw the graph of

$y =f(x)$ .

Now consider the rectangle with side lengths $a$ and $b$ . Its area is

$ab$ .

When the upper-right corner lies on the curve, so that

$b = f(a)$ ,

the curve divides the rectangle into two regions.

One has area

$\displaystyle \int_{0}^{a} f(x)\; dx$ ,

and the other has area

$\displaystyle \int_{0}^{b}g(y)\;dy$ .

Together they exactly fill the rectangle:

$\displaystyle ab = \int_{0}^{a} f(x)\;dx + \int_{0}^{b} g(y)\;dy$ .

For a general $a, b$ , the two integrals together are at least the area of the rectangle, giving

$\displaystyle ab \le \int_{0}^{a} f(x) dx + \int_{0}^{b} g(y)\; dy$ .

The picture gives us the idea.

Now let’s prove the same result algebraically.

First, consider the case

$b \ge f(a)$ .

Because $g = f^{-1}$ and $g$ is increasing, whenever $f(a) \le y \le b$ ,

$g(y) \ge g(f(a))=a.$

Therefore,

$\displaystyle \int_{f(a)}^{b} g(y)\; dy \ge \int_{f(a)}^{b} a \; dy = a(b-f(a))$ .

That is,

$\displaystyle \int_{f(a)}^{b}g(y)\;dy \ge a(b-f(a)).\quad\quad\quad(1)$

On the other hand,

$\displaystyle \int_{0}^{a} f(x)\;dx + \int_{0}^{f(a)}g(y)\;dy$

$(y=f(x),g = f^{-1}) \implies (g(y)=g(f(x)) = x, g(f(0))=0, g(f(a))= a)$

$\displaystyle = \int_{0}^{a} f(x)\;dx + \underline{\int_{0}^{a}xf'(x)\;dx}$

$\displaystyle = \int_{0}^{a} f(x)\;dx + \underline{xf(x)\bigg|_{0}^{a}-\int_{0}^{a}f(x)\;dx}$

$\displaystyle = af(a)$ .

Hence

$\displaystyle \int_{0}^{a} f(x)\;dx + \int_{0}^{f(a)}g(y)\;dy = af(a)\quad\quad\quad(*)$

It follows that

$\displaystyle \int_{0}^{a}f(x)\;dx + \underbrace{\int_{0}^{f(a)}g(y)\;dy + \int_{f(a)}^{b} g(y)\;dy}_{\int_{0}^{b}g(y)\;dy}= af(a)+\int_{f(a)}^{b}g(y)\;dy$

$\displaystyle \overset{(1)}{\ge} af(a) + a(b-f(a)) = ab$

Thus

$\boxed{ab \le \int_{0}^{a}f(x)\;dx + \int_{0}^{b}g(y)\;dy}$ .

Now consider the other case,

$b < f(a)$ .

Since $g$ is increasing,

$g(y) \le g(f(a)) = a$

whenever

$b \le y \le f(a)$ .

Therefore,

$\displaystyle \int_{b}^{f(a)}g(y)\;dy \le \int_{b}^{f(a)}a \;dy = a(f(a)-b)$

gives

$\displaystyle \int_{b}^{f(a)}g(y)\;dy \le a(f(a)-b).\quad\quad\quad(2)$

Again using (*), we obtain

$\displaystyle \int_{0}^{a}f(x)\;dx + \underbrace{\int_{0}^{f(a)} g(y)\;dy + \int_{f(a)}^{b}g(y)\;dy}_{\int_{0}^{b}g(y)\;dy}= af(a) + \int_{f(a)}^{b}g(y)\;dy$

$\displaystyle =af(a)-\int_{b}^{f(a)}g(y)\;dy$

$\displaystyle \overset{(2)}{\ge} af(a)-a(f(a)-b) = ab.$

Therefore,

$\boxed{ab \le \int_{0}^{a}f(x)\;dx + \int_{0}^{b}g(y)\;dy}$ .

So the picture and the algebra tell exactly the same story.

Now we can go back to the integral inequality that started all this.

Since it involved the powers

$x^{p-1}$

and

$y^{q-1}$ ,

let us choose

$f(x) = x^{p-1}, \qquad p>1$ .

Its inverse is

$g(y) =y^{\frac{1}{p-1}}$ .

Let $q$ satisfy

$\frac{1}{p} + \frac{1}{q} = 1$ .

Then

$q = \frac{p}{p-1}$ ,

$\frac{1}{p-1} = q-1$ .

We have

$g(y) = y^{q-1}$ .

Young’s theorem therefore gives

$\boxed{ ab \le \int_{0}^{a}x^{p-1}\;dx + \int_{0}^{b}y^{q-1}\;dy}$ .

The integral inequality from the first post has emerged as a special case of Young’s theorem.

Now we know where it comes from.

Young’s Inequality: Just Find the Minimum

October 5, 2026October 5, 2026 / Michael Xue / 1 Comment

Let $p,q>1$ satisfy $\frac1p+\frac1q=1,$

and let $a,b\ge0$ . Prove that $\frac{a^p}{p}+\frac{b^q}{q}\ge ab.$

This is Young’s inequality.

We want to prove the inequality for every pair $a,b\ge0$ .

Choose an arbitrary $b\ge0$ , and keep this $b$ fixed. We are not assigning $b$ a particular value, nor are we making any assumption about it. We simply take one arbitrary $b$ and ask:

For this $b$ , is the inequality true for every $a\ge0$ ?

If we can answer yes for an arbitrary $b$ , then the inequality is true for every $a,b\ge0$ .

With $b$ fixed, define $F(a)=\frac{a^p}{p}+\frac{b^q}{q}-ab.$

Now $F$ is a function of the single variable $a$ . Our goal is to show $F(a)\ge0.$

Differentiate: $F'(a)=a^{p-1}-b.$

The derivative vanishes when $a^{p-1}=b,$

so $a=b^{1/(p-1)}.$

From $\frac1p+\frac1q=1,$

we have $q=\frac{p}{p-1},$

and hence $q-1=\frac1{p-1}.$

Therefore the critical point is $a=b^{q-1}.$

If $a<b^{q-1}$ , then $a^{p-1}<b,$

so $F'(a)<0.$

If $a>b^{q-1}$ , then $a^{p-1}>b,$

so $F'(a)>0.$

Thus $F$ decreases up to $a=b^{q-1}$ and increases afterward. Its minimum occurs at $a=b^{q-1}.$

At this point, $a^p=b^{p(q-1)}=b^q,$

because $p(q-1)=q.$

Also, $ab=b^{q-1}b=b^q.$

Consequently, $F(a) = \frac{b^q}{p}+\frac{b^q}{q}-b^q.$

But $\frac1p+\frac1q=1,$

so $F(a)=0.$

Since $F$ has its minimum there, $F(a)\ge0.$

And because the $b$ we fixed was arbitrary, this holds for every $b\ge0$ . Therefore, for all $a,b\ge0$ , $\boxed{\frac{a^p}{p}+\frac{b^q}{q}\ge ab}.$

That’s it.

Young’s inequality, proved by fixing one variable, finding a minimum, and letting the algebra finish the job.

An Inequality Hiding in a Problem

October 2, 2026October 12, 2026 / Michael Xue / 1 Comment

Let $p,q > 1$ satisfy $\frac1p+\frac1q=1$ .

Prove: For $a,b\ge 0$ ,

$\displaystyle \int_0^a x^{p-1}\;dx+ \int_0^b x^{q-1}\;dx\ge ab$ .

At first sight, this looks like an exercise about integrals.

Let’s calculate:

$\displaystyle \int_0^a x^{p-1}\;dx=\frac{a^p}{p}, \qquad \int_0^b x^{q-1}\; dx=\frac{b^q}{q}$ .

So the problem becomes showing

$\displaystyle \frac{a^p}{p}+\frac{b^q}{q}\ge ab$ .

This looks familiar.

In fact, it is the well-known Young’s inequality.

So the original integral inequality is proved immediately.

But this raises a more interesting question:

Where did this integral inequality come from?

Can we see how it is hiding inside the original problem without carrying out the integrations?

Solving $y=x^{p-1}$ for $x$ gives

$x = y^\frac{1}{p-1}. \quad\quad\quad(1)$

And,

$\frac{1}{p} + \frac{1}{q} = 1 \implies q = \frac{p}{p-1} \implies q-1 = \frac{1}{p-1}.\quad\quad(2)$

Since

$\displaystyle \int\limits_{0}^{b} x^{q-1}\;dx = \int\limits_{0}^{b} y^{q-1}\;dy \overset{(2)}{=} \int\limits_{0}^{b} y^\frac{1}{p-1}\; dy$ ,

we have

Fig.1 $b > a^{p-1}$

Fig. 2 $b = a^{p-1}$

Fig. 3 $b < a^{p-1}$

The three cases $b>a^{p-1}, b=a^{p-1}$ , and $b<a^{p-1}$ show that the two regions together always contain the rectangle of area $ab$ . Hence, without evaluating either integral,

$\displaystyle \int_0^a x^{p-1}\; dx+\int_0^b x^{q-1} \; dx\ge ab$ .

The integral inequality did not merely happen to turn into Young’s inequality after calculation.

Young’s inequality was already there.

We have used Young’s inequality to solve the problem. We have also seen the geometry behind the integral inequality.

But we have not yet proved Young’s inequality itself.

That will be my next post.