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For $a\geq2$, $b\geq2$ and $c\geq2$, prove that

$$(a^3+b)(b^3+c)(c^3+a)\geq 125 abc.$$

My try:

First I wrote the inequality as

$$\left(a^2+\frac{b}{a}\right) \left(b^2+\frac{c}{b}\right) \left(c^2+\frac{a}{c}\right) \geq 125. $$

Then I noted that

$$a^2+\frac{b}{a}\geq a^2+\frac{2}{a}, \\ b^2+\frac{c}{b}\geq b^2+\frac{2}{b}, \\ c^2+\frac{a}{c}\geq c^2+\frac{2}{c}. $$

But I don't know how this can help me.

nonuser
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    This question is missing context or other details: Please improve the question by providing additional context, including the source of the problem and any motivation or relation to other problems in mathematics. – Carl Mummert Sep 25 '18 at 20:34
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    @CarlMummert Maybe your comment was before it was edited, but it looks fine to me. – Jakobian Sep 25 '18 at 20:43
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    @Jakobian: no source or motivation is given for the problem in the current version. The "try" is not particularly enlightening, it's just a trivial rewriting of the problem, and doesn't really add any information about the problem. – Carl Mummert Sep 25 '18 at 20:43
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    I disagree. He clearly tries to write the inequality as $f(a)f(b)f(c)\geq 125$ where $f(x)=x^2+\frac{2}{x}$. I'm not sure how many information can you give about one problem. It's probably an exercise from the book. This seems ridiculous to me – Jakobian Sep 25 '18 at 20:49
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    @Jakobian: if it is an exercise from a book, it would be easy enough to say which book! – Carl Mummert Sep 25 '18 at 20:53
  • @CarlMummert This problem is an exercise from my Brazilian book. Are you interested on it? –  Sep 25 '18 at 20:56
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    @Duarte: in this site, we are interested in posts which do more than just include a problem to solve. Yes, you should say which book you took the problem from! There is additional advice at Provide context, including what to do when you don't know how to start a problem. In general, this site is not intended merely as a place to copy problems from elsewhere without motivation - it is intended for a place to ask questions about problems. – Carl Mummert Sep 25 '18 at 20:59
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    Ho. Sorry for that then. I'm new here. Thank you for your advice. –  Sep 25 '18 at 21:02
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    It seem @CarlMummert downvote all of my answers. Why that, please? Sorry if a made mistake. – Dinesh Shankar Sep 25 '18 at 21:29
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    @Dinesh Shankar: looking at your reputation list here, only three of your answers have been downvoted today, so nobody has downvoted all of your answers. – Carl Mummert Sep 25 '18 at 22:28
  • @Carl Mummert: You're right! Sorry for that. I think I will not answer questions from new users anymore – Dinesh Shankar Sep 25 '18 at 22:50
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    I think we need to open this topic. In my opinion it's not off-topic. – Michael Rozenberg Sep 26 '18 at 00:58

3 Answers3

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$$a^3+b = {a^3\over 4}+{a^3\over 4}+{a^3\over 4}+{a^3\over 4}+b\geq 5\sqrt[5]{a^{12}b\over 2^8}$$

$$b^3+c = {b^3\over 4}+{b^3\over 4}+{b^3\over 4}+{b^3\over 4}+c\geq 5\sqrt[5]{b^{12}c\over 2^8}$$

$$c^3+a = {c^3\over 4}+{c^3\over 4}+{c^3\over 4}+{c^3\over 4}+a\geq 5\sqrt[5]{c^{12}a\over 2^8}$$

So $$\left(a^3+b\right)\left(b^3+c\right)\left(c^3+a\right) \geq 125abc\sqrt[5]{a^8b^8c^8\over 2^{24}}\geq 125abc$$

nonuser
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  • By generalized Hölder we have $$\left(a^2+\frac{b}{a}\right) \left(b^2+\frac{c}{b}\right) \left(c^2+\frac{a}{c}\right)\geq\left(\sqrt[3]{a^2b^2c^2}+\sqrt[3]{1}\right)^3\geq(\sqrt[3]{4\cdot4\cdot4}+1)^3=5^3=125.$$ – Maximilian Janisch Dec 22 '19 at 20:07
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You did 90% of the work. You can finish it by using a little bit of Calculus.

Let

$$f(x)=x^2+\frac{2}{x}.$$

Since the derivate of $f(x)$ is positive, the function $f(x)$ is increasing for $x\geq2$, and then

$$f(x)\geq f(2)=5.$$

Thereby,

$$a^2+\frac{b}{a}\geq 5, \\ b^2+\frac{c}{b}\geq 5,\\ c^2+\frac{a}{c}\geq 5,$$

which implies that

$$ \left(a^2+\frac{b}{a}\right) \left(b^2+\frac{c}{b}\right) \left(c^2+\frac{a}{c}\right) \geq 125.$$

Dinesh Shankar
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    It is not necessary to use calculus to prove monotonicity of $f$ on $[2,\infty)$, though, but good solution nonetheless. – Batominovski Sep 25 '18 at 21:17
  • This is easiest, I'd say. You can even just do this before the division with $ a^3 + b \geq 10 $, directly using the inequalities. – theREALyumdub Sep 25 '18 at 23:00
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Another way.

By AM-GM for all $x\geq2$ we obtain: $$x^3+16=x^3+8+8\geq3\sqrt[3]{x^3\cdot8^2}=12x.$$ Thus, $$\prod_{cyc}(a^3+b)\geq\prod_{cyc}(12a-16+b)=\prod_{cyc}(5a+7a+b-16)\geq\prod_{cyc}5a=125abc.$$

Michael Rozenberg
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