Problem 12: Vasc inequality

by henderson, May 18, 2016, 8:23 PM

We don't even need the reals $a$ , $b$ , $c$ to be positive!
$\bf\color{red}Problem\ 12\$ If $a$ , $b$ , $c$ are real numbers, then prove that
$a^2 + b^2 + c^2 \geq \sqrt {3\left(a^3b + b^3c + c^3a\right)}$ .
$\bf\color{red}Solution\$ Recently, a person, who did not want his name to be mentioned - I will subsequently refer to him under the pseudonym Stefan V. -, proved this inequality by the following identity:

$\left(a^2 + b^2 + c^2\right)^2 - 3\left(a^3b + b^3c + c^3a\right)$
$= \frac {1}{2}\left( \left( a^{2} - 2ab + bc - c^{2} + ca\right) ^2 + \left(b^{2} - 2bc + ca - a^{2} + ab\right)^2 + \left( c^{2} - 2ca + ab - b^{2} + bc\right)^2 \right)$
$\geq 0$ .

This actually shows that the inequality $\left(a^2 + b^2 + c^2\right)^2\geq 3\left(a^3b + b^3c + c^3a\right)$ holds for any three real (not necessarily positive!) numbers a, b, c.

Also, Stefan noted that the inequality becomes an equality in the case when a = b = c and in the case when $a: b: c = \sin^2\frac {4\pi}{7}: \sin^2\frac {2\pi}{7}: \sin^2\frac {\pi}{7}$ and in the cyclic permutations of this case. This strange equality case is what makes the inequality so difficult to prove.

These are some few of Stefan's investigations concerning this inequality. Let me add the mine. In post #10, Vasc established his inequality $\left(a^2 + b^2 + c^2\right)^2\geq 3\left(a^3b + b^3c + c^3a\right)$ using the identity

$4\left(\left(a^2 + b^2 + c^2\right) - \left(bc + ca + ab\right)\right)\left(\left(a^2 + b^2 + c^2\right)^2 - 3\left(a^3b + b^3c + c^3a\right)\right)$
$= \left(\left(a^3 + b^3 + c^3\right) - 5\left(a^2b + b^2c + c^2a\right) + 4\left(b^2a + c^2b + a^2c\right)\right)^2$
$+ 3\left(\left(a^3 + b^3 + c^3\right) - \left(a^2b + b^2c + c^2a\right) - 2\left(b^2a + c^2b + a^2c\right) + 6abc\right)^2$ .

Actually, this may look a miracle, but there is a very natural way to find this identity. In fact, we consider the function

$g\left(a;\;b;\;c\right) = \left(a^2 + b^2 + c^2\right)^2 - 3\left(a^3b + b^3c + c^3a\right)$

over all triples $\left(a;\;b;\;c\right)\in\mathbb{R}^3$ . We want to show that this function satisfies $g\left(a;\;b;\;c\right)\geq 0$ for any three reals a, b, c. Well, fix a triple (a; b; c) and translate it by some real number d; in other words, consider the triple (a + d; b + d; c + d). For which $d\in\mathbb{R}$ will the value $g\left(a + d;\;b + d;\;c + d\right)$ be minimal? Well, minimizing $g\left(a + d;\;b + d;\;c + d\right)$ is equivalent to minimizing $g\left(a + d;\;b + d;\;c + d\right) - g\left(a;\;b;\;c\right)$ (since (a; b; c) is fixed), but

$g\left(a + d;\;b + d;\;c + d\right) - g\left(a;\;b;\;c\right)$
$= d^2\left(\left(a^2 + b^2 + c^2\right) - \left(bc + ca + ab\right)\right)$
$+ d\left(\left(a^3 + b^3 + c^3\right) - 5\left(a^2b + b^2c + c^2a\right) + 4\left(b^2a + c^2b + a^2c\right)\right)$ ,

so that we have to minimize a quadratic function, what is canonical, and it comes out that the minimum is achieved for

$d = - \frac {\left(a^3 + b^3 + c^3\right) - 5\left(a^2b + b^2c + c^2a\right) + 4\left(b^2a + c^2b + a^2c\right)}{2\left(\left(a^2 + b^2 + c^2\right) - \left(bc + ca + ab\right)\right)}$ .

So this is the value of d such that $g\left(a + d;\;b + d;\;c + d\right)$ is minimal. Hence, for this value of d, we have $g\left(a;\;b;\;c\right)\geq g\left(a + d;\;b + d;\;c + d\right)$ . Thus, in order to prove that $g\left(a;\;b;\;c\right)\geq 0$ , it will be enough to show that $g\left(a + d;\;b + d;\;c + d\right)\geq 0$ . But, armed with the formula

$d = - \frac {\left(a^3 + b^3 + c^3\right) - 5\left(a^2b + b^2c + c^2a\right) + 4\left(b^2a + c^2b + a^2c\right)}{2\left(\left(a^2 + b^2 + c^2\right) - \left(bc + ca + ab\right)\right)}$

and with a computer algebra system or a sufficient patience, we find that

$g\left(a + d;\;b + d;\;c + d\right) = \frac {3\left(\left(a^3 + b^3 + c^3\right) - \left(a^2b + b^2c + c^2a\right) - 2\left(b^2a + c^2b + a^2c\right) + 6abc\right)^2}{4\left(\left(a^2 + b^2 + c^2\right) - \left(bc + ca + ab\right)\right)}$ ,

what is incontestably $\geq 0$ . So we have proven the inequality. Now, writing

$g\left(a;\;b;\;c\right) = g\left(a + d;\;b + d;\;c + d\right) - \left(g\left(a + d;\;b + d;\;c + d\right) - g\left(a;\;b;\;c\right)\right)$

and performing the necessary calculations, we arrive at Vasc's mystic identity.

This post has been edited 3 times. Last edited by henderson, Sep 12, 2016, 2:33 PM

Vasc inequality

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$\bf\color{green}Another\ proof\$ If $x,\,y,\,z$ are real numbers then
$(x+y+z)^2 \geq 3(xy+yz+zx). \quad (1)$ Replacing $x = a^2+b(c-a),\,y = b^2+c(a-b),\,z = c^2+a(b-c)$ inequality $(1)$ becomes
$(a^2+b^2+c^2)^2\geq 3(a^3b+b^3c+c^3a).$ We are done.

This post has been edited 2 times. Last edited by henderson, May 31, 2016, 6:12 PM

by henderson, May 18, 2016, 8:37 PM

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Problem 12: Vasc inequality

by henderson, May 18, 2016, 8:23 PM

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