# Does the Young's inequality $x^2+y^2 \geq 2xy$ holds for negative x or y?

I have noticed in the literature that Young's inequality $x^2+y^2 \geq 2xy$ is mentioned considering non-negative $x$ and $y$. But considering various cases, I have found that the inequality holds for negative non zero $x$ and $y$. Does the Young's inequality holds for negative $x$ or $y$ in general?

Case:1 $x=-1$ and $y=1$. $\implies1+1\geq-1$ (Inequality holds).

Case:2 $x=1$ and $y=-1$. $\implies 1+1\geq-1$ (Inequality holds).

Case:3 $x=-1$ and $y=-1$. $\implies 1+1 \geq 1$ (Inequality holds)

• No, it's always stated for all $x,y \in \mathbb R$. It is always true that $(x-y)^2 \ge 0$. – Cauchy Aug 28 '17 at 7:06
• In yourr three cases, the numbers on the right should be $-2,-2,2$ instead of $-1,-1,1$. But yeah, the inequality still holds. – 5xum Aug 28 '17 at 7:16

The Young inequality it's the following.

Let $x$, $y$ be non-negative numbers, $p$ and $q$ be positive numbers such that $\frac{1}{p}+\frac{1}{q}=1$. Prove that: $$\frac{x^p}{p}+\frac{y^q}{q}\geq xy.$$ Indeed, for $p=q=2$ we get $x^2+y^2\geq2xy$,

but the last inequality is true for all reals $x$ and $y$ because it's just $(x-y)^2\geq0$.

But in the general case it's wrong.

For example. Try $p=\frac{3}{2}$ and $q=3$.

Hence, we obtain $$\frac{2}{3}x^{\frac{3}{2}}+\frac{1}{3}y^3\geq xy ,$$ which is obviously wrong for $x=-1$.

• But for the general inequality you stated, it can't hold for negative $x,y$ since those powers are not even necessarily well-defined. – wythagoras Aug 28 '17 at 7:18
• @wythagoras Yes of course! – Michael Rozenberg Aug 28 '17 at 7:18

The inequality $x^2+y^2\geq 2xy$ holds for all $x,y\in \Bbb R$; indeed, it is equivalent to $(x-y)^2\geq 0$, which always holds when $x-y\in \Bbb R$. Note that changing the sign of $x$ or $y$ only change the RHS, so if the inequality holds for nonnegative numbers it must hold for all reals.

The reason you've seen it stated for nonnegative $x,y$ is probably that a lot of inequalities are only valid for nonnegative numbers, so it is "usual" to add that hypothesis, even though here it is not necessary.

As a side note, I've never seen this inequality being called "Young's inequality" : for me that name applies to the more general inequality $$\frac{x^p}{p}+\frac{y^{q}}{q}\geq xy,$$ where $\frac{1}{p}+\frac{1}{q}=1$ (your case is the one where $p=q=2$); this one is in general only true for $x,y\geq 0$, since otherwise $x^p$ and $y^q$ wouldn't be defined.

• I hadn't seen Michael's answer before I edited mine, it's a bit redundant now... – Arnaud D. Aug 28 '17 at 7:21