The Tangent-line Origin of Young’s Inequality

Young’s inequality can look like a trick the first time you meet it. But the real source of the inequality is not algebraic cleverness. It comes from a geometric fact about a concave function.

For \(f(x)=x^{1/p}\) with \(p>1\), the derivative is

$$ f'(x)=\frac{1}{p}x^{1/p-1}. $$

Notice that this function is concave. It bends downward. Concavity gives us a built-in comparison tool: a concave function always lies below its tangent lines. Because tangent lines are linear, and linear expressions are much easier to work with than curved ones we can replace a curved expression with a linear upper bound.

At \(x=1\), we have

$$ f(1)=1 \qquad\text{and}\qquad f'(1)=\frac{1}{p}. $$

We choose $𝑥=1$ because the arithmetic becomes especially clean there. The function value is $1$, the slope is $1/𝑝$, and the resulting line will contain the coefficients that later become $1/𝑝$ and $1/𝑞$.

The tangent line at \(x=a\) has the general form

$$ L(x)=f(a)+f'(a)(x-a). $$

Taking \(a=1\), this becomes

$$ L(x)=f(1)+f'(1)(x-1). $$

Substituting the values gives

$$ L(x)=1+\frac{1}{p}(x-1). $$

Expanding, we get

$$ L(x)=1+\frac{1}{p}x-\frac{1}{p} =\frac{1}{p}x+\left(1-\frac{1}{p}\right). $$

If \(\frac{1}{p}+\frac{1}{q}=1\), then \(1-\frac{1}{p}=\frac{1}{q}\), so

$$ L(x)=\frac{1}{p}x+\frac{1}{q}. $$

We introduce $𝑞$ here because Young’s inequality is written in terms of the conjugate exponents $𝑝$ and $𝑞$. This step reveals that the coefficient $1/𝑞$ is not inserted by hand. It comes directly from the tangent-line geometry.

Since \(f(x)=x^{1/p}\) is concave on \([0,\infty)\), its graph lies below its tangent line at \(x=1\). Therefore, for all \(x\ge 0\),

$$ x^{1/p}\le 1+\frac{1}{p}(x-1)=\frac{1}{p}x+\frac{1}{q} $$

A deeper insight into why we chose $x^{1/p}$ was something like this: If I want to control a nonlinear quantity by a linear one, I should look for a concave function whose tangent line produces the right coefficients. That is a very general mathematical move: identify the difficult nonlinear thing, find a geometric or structural property that linearizes it, and then choose the function that matches the exponents already present in the problem.

Links to this note:
Young’s Inequality for Products