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 $a,b\in\mathbb R$?

Question

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 !

Answer 1

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.

Answer 2

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.