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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?

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    $\begingroup$ The statement is precisely that the function $f(x)=x-(x^{3}-cx^{2})^{1/3}$ is a decreasing function on $(0,\infty)$. Use derivative. $\endgroup$ – Kabo Murphy Oct 19 '18 at 9:07
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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 $c\geq\frac{9ab(a^2-ab+b^2)}{(a+b)^3}.$

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  • $\begingroup$ @Kyky See please your first line of your post. You say there about solving. In any case I proved that your inequality is wrong. See please better my post. $\endgroup$ – Michael Rozenberg Oct 19 '18 at 10:36
  • $\begingroup$ I just noted a mistake from your post: $a+b-c>0$ does not always equal $a^3+b^3-c^3>0$, an example is $a,b=2, c=3$ Also, may I apologize for not posting this earlier. $\endgroup$ – Kyky Oct 19 '18 at 11:01
  • $\begingroup$ @Kyky I used $a-b-c>0$ equal to $a^3-b^3-c^3-3abc>0$, when $(a+b)^2+(a+c)^2+(b-c)^2\neq0.$ See please better my post. $\endgroup$ – Michael Rozenberg Oct 19 '18 at 11:17
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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 $(0, \infty)$.

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