Purely "algebraic" proof of Young's Inequality

Young's inequality states that if $a, b \geq 0$, $p, q > 0$, and $\frac{1}{p} + \frac{1}{q} = 1$, then $$ab\leq \frac{a^p}{p} + \frac{b^q}{q}$$ (with equality only when $a^p = b^q$). Back when I was in my first course in real analysis, I was assigned this as homework, but I couldn't figure it out. I kept trying to manipulate the expressions algebraically, and I couldn't get anywhere. But every proof that I've seen since uses calculus in some way to prove this. For example, a common proof is based on this proof without words and integration. The proof on Wikipedia uses the fact that $\log$ is concave, which I believe requires the analytic definition of the logarithm to prove (correct me if I'm wrong).

Can this be proven using just algebraic manipulations? I know that that is a somewhat vague question, because "algebraic" is not well-defined, but I'm not sure how to make it more rigorous. But for example, the proof when $p = q = 2$ is something I would consider to be "purely algebraic":

$$0 \leq (a - b)^2 = a^2 + b^2 - 2ab,$$ so $$ab \leq \frac{a^2}{2} + \frac{b^2}{2}.$$

Solution 1:

This proof is from "Mathematical Toolchest" published by the Australian Mathematics Trust (image).

Example. If $p$ and $q$ are positive rationals such that $\frac1p + \frac1q = 1$, then for positive $x$ and $y$ $$\frac{x^p}p + \frac{y^q}q \ge xy.$$

Since $\frac1p + \frac1q = 1$, we can write $p = \frac{m+n}m$, $q = \frac{m+n}n$ where $m$ and $n$ are positive integers. Write $x = a^{1/p}$, $y = b^{1/q}$. Then $$\frac{x^p}p + \frac{y^q}q = \frac a{\frac{m+n}m} + \frac b{\frac{m+n}n} = \frac{ma + nb}{m + n}.$$

However, by the AM–GM inequality, $$\frac{ma + nb}{m + n} \ge (a^m \cdot b^n)^{\frac1{m+n}} = a^{\frac1p} b^{\frac1q} = xy,$$ and thus $$\frac{x^p}p + \frac{y^q}q \ge xy.$$

Solution 2:

Yes, at least for rational $p, q$. There is a general statement here, which can be summarized as follows:

Every polynomial inequality is a consequence of the trivial inequality $x^2 \ge 0$.

In more detail, to prove Young's inequality for general $p, q$, it suffices by a continuity argument to prove it for $p, q$ rational. By making an appropriate substitution of the form $a = x^n, b = y^m$ where $n, m$ are even integers, Young's inequality for a fixed choice of rational $p, q$ becomes equivalent to the statement that a certain polynomial with real coefficients always takes on non-negative real values when fed real inputs.

By Artin's solution to Hilbert's 17th problem, a polynomial with this property is a sum of squares of rational functions. An expression of this polynomial as a sum of squares of rational functions constitutes a proof of the corresponding case of Young's inequality from repeated application of the trivial inequality.

I don't see how you could avoid analysis for irrational $p, q$, since you can't even define the relevant functions without analysis.

Solution 3:

With $\dfrac1p+\dfrac1q=1$, $u=x^p$, $v=y^q$, and $1+pt=u/v$, the following are equivalent: $$ \begin{align} xy&\le\frac{x^p}{p}+\frac{y^q}{q}\\ u^{1/p}v^{1/q}&\le\frac{u}{p}+\frac{v}{q}\\ (u/v)^{1/p}&\le\frac{u/v}{p}+\frac{1}{q}\\ (1+pt)^{1/p}&\le\frac{1+pt}{p}+\frac{1}{q}\\ 1+pt&\le(1+t)^p\tag{1} \end{align} $$ Where $(1)$ is the rational version of the Bernoulli inequality, proven below.

Bernoulli's Inequality for Rational Exponents

Using the integral version of the Bernoulli Inequality, proven at the end of this answer, we get that for $x\gt-n$, $$ \begin{align} \frac{\left(1+\frac{x}{n+1}\right)^{n+1}}{\left(1+\frac{x}{n}\right)^n} &=\left(\frac{(n+x+1)n}{(n+1)(n+x)}\right)^{n+1}\frac{n+x}{n}\\ &=\left(1-\frac{x}{(n+1)(n+x)}\right)^{n+1}\frac1{1-\frac{x}{n+x}}\\ &\ge\left(1-\frac{x}{n+x}\right)\frac1{1-\frac{x}{n+x}}\\[8pt] &=1\\[8pt] \left(1+\frac{x}{n+1}\right)^{n+1} &\ge\left(1+\frac{x}{n}\right)^n\tag{2} \end{align} $$ where the inequality is strict if $x\ne0$ and $n\ge1$.

Applying induction with $(2)$, we get that for $x\gt-m$ and integers $n\ge m$, $$ \left(1+\frac{x}{n}\right)^n\ge\left(1+\frac{x}{m}\right)^m\tag{3} $$ Letting $t=\frac{x}{n}\gt-\frac{m}{n}$ and taking $m^\text{th}$ roots gives $$ \left(1+t\right)^{n/m}\ge1+\frac{n}{m}t\tag{4} $$ Note that $(4)$ is trivially true for $-1\le t\le-\frac{m}{n}$. Thus, for all $t\ge-1$ and rational $p\ge1$, $$ \left(1+t\right)^p\ge1+pt\tag{5} $$ where the inequality is strict if $t\ne0$ and $p\gt1$.

Negative Exponents

As shown in $(2)$, both $\left(1+\frac xn\right)^n$ and $\left(1-\frac xn\right)^n$ are increasing in $n$, as long as $|x|\le n$. Thus, for $m=\max(k,n)$, $$ \begin{align} \left(1+\frac xk\right)^k\left(1-\frac xn\right)^n &\le\left(1+\frac xm\right)^m\left(1-\frac xm\right)^m\\ &=\left(1-\frac{x^2}{m^2}\right)^m\\[3pt] &\le1\tag6 \end{align} $$ Thus, substituting $x\mapsto kx$, and taking $n^\text{th}$ roots, we get $$ (1+x)^{k/n}\left(1-\tfrac knx\right)\le1\tag7 $$ Finally, setting $p=\tfrac kn$ yields $$ 1-px\le(1+x)^{-p}\tag8 $$ where the inequality is strict when $x\ne0$ and $p\gt0$.

Solution 4:

It might be essentially easier to write $x=a^p,y=b^q$ and $t=\frac 1 p$. Then you want to prove the inequality:

$$x^{t}y^{1-t}\leq tx + (1-t)y$$

for $0<t<1$ and with equality only when $x=y$.

First, we prove the general AM-GM for any $2^k$ variables. The case $k=1$ is the obvious case:

$$(x+y)^2-4xy = (x-y)^2\geq 0$$

With equality only when $x=y$.

Now assume we have proven AM-GM for $n$ terms, we will prove it for $2n$ terms.

If $x_1,...,x_{2n}$ are real, assume they are in linear order. Then:

$$\begin{align}\sqrt[2n]{x_1...x_{2n}} &= \sqrt{\sqrt[n]{x_1...x_n}\sqrt[n]{x_{n+1}...x_{2n}}}\\ &\leq \frac{1}{2}\sqrt[n]{x_1...x_n}+\frac{1}{2}\sqrt[n]{x_{n+1}...x_{2n}} \\ &\leq \frac{1}{2n}(x_1+...+x_n)+\frac{1}{2n}(x_{n+1}+...+x_{2n}) \end{align}$$

Since we linearly ordered them, the equality holds only if all the $x_i$ are equal. (Check yourself here.)

So, by induction, the AM/GM applies to any set of $2^k$ variables. If $t=r/2^k$, we can choose $x_1=x_2=..=x_r=x$ and $x_{r+1}=...=x_s=y$. Then $$\sqrt[2^k]{x^ry^{2^k-r}}\leq \frac{r}{2^k} x + \frac{2^k-r}{2^k} y$$

Which is just $$x^ty^{1-t}\leq tx + (1-t)y$$

(That was just Ben's argument above, but restricted to the $2^k$ case.)

Since the set of $t$ of the form $r/2^k$, $r,k\in\mathbb Z$ is dense in $(0,1)$, you have this inequality everywhere (although density does not obviously let you conclude that equality only occurs when $x=y$ for the arbitrary real $t$.)