Prove inequality if A and B are real positive numbers

Question

I am trying to solve the following problem but I don't know how to start. I am trying to prove that if $a$ and $b$ are real, positive numbers then the following equation is true:$$\left (\frac{a+b}{2}\right )^3+\left (\frac{a-b}{2}\right )^3\leq \frac{a^3+b^3}{2}.$$Any help/tips are appreciated.

Answer 1

An easy computation shows

$$\frac{a^3+b^3}{2}-\left(\frac{a+b}{2}\right)^3-\left(\frac{a-b}{2}\right)^3 = \frac{a^3-3ab^2+2b^3}{4},$$

so we just have to show

$$a^3-3ab^2+2b^3 \geq 0 \quad \quad \quad (*)$$

for $a,b > 0$.

Let $t:=\frac{a}{b}$. Dividing inequality $(*)$ by $b^3$, we get $(*)$ is equivalent to

$$t^3-3t+2 \geq 0. \quad \quad \quad (**)$$

Here we are using the fact that $b > 0$.

Note $t^3-3t+2 = (t-1)^2(t+2)$, so $(**)$ holds because $t=\frac{a}{b} > 0$.