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Is it true that

$$ (\alpha+\beta+\gamma)(\alpha\beta\gamma-\alpha^3)\leq (\frac{\alpha\beta+\alpha\gamma+\beta\gamma}{2})^2, $$

for all real numbers $\alpha,\beta,\gamma$?

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    Posts the merely state a question, like this one, are discouraged on this site. Seehttp://meta.math.stackexchange.com/questions/9959/how-to-ask-a-good-question for some suggestions on how to write a better post. The best posts have exposition that explain the source or inspiration of the problem, its interest, and its relationships with other problems, in a way that makes the problem compelling. – 2017-01-28

3 Answers 3

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It's $(2\alpha^2+\alpha\beta+\alpha\gamma-\beta\gamma)^2\geq0$

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    How did you derive that? – 2017-01-25
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    @S.C.B. After full expanding I used the gland inside the skull. – 2017-01-25
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    What method is that? – 2017-01-25
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    @S.C.B A full expanding gives $4a^4+a^2b^2+a^2c^2+b^2c^2+4a^3b+4a^3c-2a^2bc-2b^2ac-2c^2ab\geq0$, which gives the answer. – 2017-01-25
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It is equivalent to $$1/4\, \left( 2\,{\alpha}^{2}+\alpha\,\beta+\alpha\,\gamma-\beta\, \gamma \right) ^{2} \geq 0$$ which is true.

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Not an answer but a simple remark ...

This result implies that given any equation of the form

$$x^3+ax^2+bx+c=0$$

where $a,b,c$ are any real numbers, and if $\alpha$ denotes a real root (and such a root exists !), then we have :

$$a(c+\alpha^3)\le\left(\frac{b}{2}\right)^2$$

In particular, if $a>0$ then :

$$\alpha\le\left(\frac{b^2-4ac}{4a}\right)^{1/3}$$

I was not aware of this result ...

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    @ Adren. Well done!, I derived the inequality from the cubic polynomial. – 2017-01-25