Young's inequality (for products) and the AM-GM inequality: random musings

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 Young's inequality (for products) is sometimes called the generalised AM-GM (arithmetic mean -geometric mean) inequality.

Let $a$ and $b$ be non-negative real numbers. Then $(ab)^{1/2} \leq (a+b)/2$, with equality if and only if $a=b$. This can be seen easily by multiplying both sides by $2$, then squaring both sides and noting that $(a+b)^2=(a-b)^2+4ab$. This is the special case of the AM-GM inequality for two non-negative real numbers (and the result generalises to any finite number of non-negative real numbers). However it is also the special case of Young's inequality when $p=q=2$.

Young's inequality states that for non-negative real numbers $x$ and $y$ and positive real numbers $p$ and $q$ satisfying $1/p + 1/q = 1$, we have $xy \leq x^p/p + y^q/q$, with equality if and only if $x^p=y^q$. When $p=q=2$ this says that $xy \leq (x^2+y^2)/2$ with equality if and only if $x^2=y^2$, which is the same as the version of AM-GM above above when you set $a=x^2$ and $b=y^2$.

Both AM-GM and Young's inequality can be proved in many ways. But (for some reason) I found myself wondering whether Young's inequality can be deduced from the (full version of the) AM-GM inequality.

I noted that, by continuity, it is sufficient to prove Young's inequality for positive rational numbers $p$ and $q$ such that $1/p + 1/q=1$, and that we can then find positive integers $m$ and $n$ such that $p=(m+n)/m$ and $q=(m+n)/n$.

Next, let $a$ and $b$ be non-negative real numbers, and note that $a^m b^n$ can be thought of as a product of $m+n$ non-negative real numbers: $m$ copies of $a$ and $n$ copies of $b$. Applying the full version of the AM-GM inequality to that gives us

\[(a^m b^n)^{1/(m+n)} \leq (m a + n b)/(m+n)\,,\]

i.e.,

\[a^{1/p} b^{1/q} \leq a/p + b/q\,,\]

with equality if and only if $a=b$.

Finally, let $x$ and $y$ be non-negative real numbers, and apply the above to $a=x^p$ and $b=y^q$ to obtain

\[xy \leq x^p/p + y^q/q\,,\]
with equality if and only if $x^p=y^q$.

This proves Young's inequality for rational $p$ and $q$ as above, and the general case follows (with a little care) by continuity. 

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