Intereting Posts

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Complex Exponents

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 $$ab \leq \frac{a^2}{2} + \frac{b^2}{2}.$$

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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 $$\frac{x^p}p + \frac{y^q}q \ge xy.$$

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.

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.

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 $p\gt1$.

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 $t$.)

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