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 !

The first inequality is equivalent to $a^2+b^2\leq 2( a^2+b^2)+2ab$, i.e., to $0\leq (a+b)^2$, which obviously is true.
The second is equivalent to $2( a^2+b^2+ab)\leq 3( a^2+b^2)$, i.e., to $0\leq (a-b)^2$ which again is obviously true.
