Does the Young's inequality $x^2+ y^2\geq 2xy$ hold for negatives $x$ 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 hold for negatives $x$ or $y$ in general ?
Case 1 : $x=-1, y=1\implies1+1\geq-1$ (Inequality holds).
Case 2 : $x=1, y=-1\implies 1+1\geq-1$ (Inequality holds).
Case 3 : $x=-1, y=-1\implies 1+1 \geq 1$ (Inequality holds)
 A: 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 $x=-1$.
A: 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.
