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

  • $\begingroup$ By the way, I wasn't quite sure how to properly tag this question. Apparently proof is not allowed. $\endgroup$
    – asmeurer
    Dec 16, 2012 at 7:22
  • $\begingroup$ I think fundamentally, there is no algebraic definition of $x^p$ in general when $p,q$ are real. But Ben's proof is good for when $p,q$ are rational. $\endgroup$ Dec 16, 2012 at 8:14
  • $\begingroup$ @ThomasAndrews yes, I know that, but it still satisfies certain algebraic properties, which I would take for granted. For example, conditions on $x$ or $p$ for $x^p < x$ to hold. $\endgroup$
    – asmeurer
    Dec 16, 2012 at 8:25
  • $\begingroup$ @asmeurer I've retagged. The original tags you chose are a bit too specialized. The closest I can find to a tag that match your interpretation of "algebraic" is (algebra-precalculus), where the tag-wiki explicitly mentions symbolic manipulations, which is probably similar to what you had in mind. $\endgroup$ Dec 17, 2012 at 16:41

4 Answers 4


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

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    $\begingroup$ That assumes $p,q$ are rational, of course. $\endgroup$ Dec 16, 2012 at 8:08
  • $\begingroup$ Cool. I didn't know of this inequality at the time, so it wouldn't have occurred to me. Apparently Young's Inequality actually is a special case of the weighted AM-GM inequality. $\endgroup$
    – asmeurer
    Dec 16, 2012 at 8:19
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    $\begingroup$ @robjohn What do you mean? All that's used is plain old AM–GM? $\endgroup$
    – Ben
    Jan 27, 2014 at 5:25
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    $\begingroup$ @Ben: it depends on what you consider the "plain old AM-GM" inequality: the basic AM-GM is $\sqrt{xy}\le\frac{x+y}{2}$. The version $x^\alpha y^{1-\alpha}\le\alpha x+(1-\alpha)y$ with $0\le\alpha\le1$ can be proven for dyadic $\alpha$ by induction on the basic case. Other $\alpha\in\mathbb{Q}$ can be achieved separately using algebra, or after establishing the inequality for all $\alpha\in\mathbb{R}$ using analysis. $\endgroup$
    – robjohn
    Jan 27, 2014 at 14:21
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    $\begingroup$ @robjohn I see, thank you. I would consider plain AM–GM to be $\frac{x_1 + x_2 + \cdots + x_n}{n} \ge \sqrt[n]{x_1x_2 \cdots x_n}$ which can be shown with backward induction. $\endgroup$
    – Ben
    Jan 27, 2014 at 22:49

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.

  • $\begingroup$ I don't understand your statement "Every polynomial inequality is a consequence of the trivial inequality $x^2 \geq 0$." This seems to contradict the following statement in the wiki article you have cited. "The formulation of the question takes into account that there are polynomials, for example $$x^6 + x^4 y^2 + x^2 y^4 - 3x^2 y^2 z^2$$ which are non-negative over reals and yet which cannot be represented as a sum of squares of other polynomials, as Hilbert had shown in $1888$ but without giving an example: the first explicit example was found by Motzkin in $1966$." $\endgroup$
    – user17762
    Dec 16, 2012 at 9:02
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    $\begingroup$ @asmeuer: it is not necessary to find them to prove that they exist, which Artin showed. $\endgroup$ Dec 17, 2012 at 3:42
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    $\begingroup$ I thought the point here was to find them to prove the inequality. Artin's theorem tells us that our search is a reasonable direction to take (and that purely algebraic proofs are possible, at least modulo choosing rational exponents only). $\endgroup$
    – asmeurer
    Dec 17, 2012 at 20:22
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    $\begingroup$ On this topic, see also andrescaicedo.wordpress.com/2008/11/11/275-positive-polynomials $\endgroup$ Dec 23, 2012 at 6:29
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    $\begingroup$ @AndresCaicedo excellent blog post! $\endgroup$
    – asmeurer
    Dec 24, 2012 at 4:46

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 $p\gt0$.


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 $t$.)

  • $\begingroup$ This does prove the inequality for the case of $t:t\,2^k\in\mathbb{Z}$. (+1) $\endgroup$
    – robjohn
    Mar 11, 2013 at 6:10
  • $\begingroup$ Yeah, to get for all rationals, you need the full AM/GM. And to get for all real $t$, you need continuity and the density of the rationals in the reals. @robjohn $\endgroup$ Mar 11, 2013 at 6:13
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    $\begingroup$ I think my proof works for all rationals. It uses a pretty simply proven extension of Bernoulli's Inequality. $\endgroup$
    – robjohn
    Mar 11, 2013 at 6:19

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