Is $\dfrac{a^2+b^2}{6}\leq \dfrac{a^2+b^2}{3}+\dfrac{ab}{3}\leq \dfrac{a^2+b^2}{2}$ for any $a,b\in\mathbb R$? Is $\dfrac{a^2+b^2}{6}\leq \dfrac{a^2+b^2}{3}+\dfrac{ab}{3}\leq \dfrac{a^2+b^2}{2}$ for any $a,b\in\mathbb R$ ?
For $a$ and $b$ are both positive or both negative,I proved this.
But I am not able to prove for all $a,b\in\mathbb R$ ?
Please help me !

Suppose $a \ge 0$
and $b < 0$.
Writing $-b$ for $b$,
this becomes
$\dfrac{a^2+b^2}{6}\leq \dfrac{a^2+b^2}{3}-\dfrac{ab}{3}\leq \dfrac{a^2+b^2}{2}
$
with $a \ge 0, b > 0$.
The right-hand inequality
is obviously true.
The left-hand one is
$ab/3 \le (a^2+b^2)/6$
or
$2ab \le a^2+b^2$
or
$0 \le (a-b)^2$
which is true.
