Prove or disprove inequality with 3 variables

Question

I was trying to solve the inequality $$a-\sqrt[3]{a^3-c\cdot a^2}<b-\sqrt[3]{b^3-c\cdot b^2}$$ where $a>b>0$ and $c>0$. I managed to pack the part inside the cube root: $$a-\sqrt[3]{a^2(a-c)}<b-\sqrt[3]{b^2(b-c)}$$ but I'm stuck after that. Can anybody help prove/disprove the inequality?

Answer 1

We need to solve $$\sqrt[3]{a^2(a-c)}-\sqrt[3]{b^2(b-c)}-(a-b)>0$$ or since $$a^2(a-c)=-b^2(b-c)=-(a-b)^3$$ gives $$\frac{a^2b^2}{a^2+b^2}+(a-b)^2=0,$$ which is impossible, we need to solve $$a^2(a-c)-b^2(b-c)-(a-b)^3-3(a-b)\sqrt[3]{a^2b^2(a-c)(b-c)}>0$$ or $$3ab-ac-bc>3\sqrt[3]{a^2b^2(a-c)(b-c)}$$ or $$c<\frac{9ab(a^2-ab+b^2)}{(a+b)^3}.$$ By the way, if you need to prove this inequality then it's wrong for $c\geq\frac{9ab(a^2-ab+b^2)}{(a+b)^3}.$

Answer 2

For fixed $c > 0$, the function $f(x) = x-\sqrt[3]{x^3-c x^2}$ satisfies

• $f(0) = 0$,
• $f(c) = c$,
• $\lim_{x \to \infty} f(x) = \frac c3$,

therefore $f$ is not monotonic on $(0, \infty)$.