Three variable inequality with sum of one

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Three variable inequality with sum of one

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Source: 239 2012 J4

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#1 h Jul 30, 2020, 3:17 AM • 2 Y

Y by pasionfruit, Mango247

For positive real numbers $a$ , $b$ , and $c$ with $a+b+c=1$ , prove that:
$(a-b)^2 + (b-c)^2 + (c-a)^2 \geq \frac{1-27abc}{2}.$

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#2 h Jul 30, 2020, 4:24 AM

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SinaQane wrote:

For positive real numbers $a$ , $b$ , and $c$ with $a+b+c=1$ , prove that:
$(a-b)^2 + (b-c)^2 + (c-a)^2 \geq \frac{1-27abc}{2}.$

By well-know identity: $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ We have: $a^3+b^3+c^3-3abc=(a^2+b^2+c^2-ab-bc-ca)$ Then: $(a-b)^2 + (b-c)^2 + (c-a)^2=2(a^2+b^2+c^2-ab-bc-ca)=2(a^3+b^3+c^3-3abc)\ge \frac{1-27abc}{2}$
P.S. watch @below to see explanation of last step

This post has been edited 4 times. Last edited by CROWmatician, Jul 30, 2020, 10:03 AM
Reason: .

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MortemEtInteritum

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MortemEtInteritum

#3 h Jul 30, 2020, 4:45 AM

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CROWmatician wrote:

SinaQane wrote:

For positive real numbers $a$ , $b$ , and $c$ with $a+b+c=1$ , prove that:
$(a-b)^2 + (b-c)^2 + (c-a)^2 \geq \frac{1-27abc}{2}.$

By well-know identity: $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ We have: $a^3+b^3+c^3-3abc=(a^2+b^2+c^2-ab-bc-ca)$ Then: $(a-b)^2 + (b-c)^2 + (c-a)^2=2(a^2+b^2+c^2-ab-bc-ca)=2(a^3+b^3+c^3-3abc)\ge 6-6abc\ge \frac{1-27abc}{2}$
P.S. You can prove that $6-6abc\ge \frac{1-27abc}{2}$ , by moving terms and noting that $15abc>-11$ .

$2(a^3+b^3+c^3-3abc)\ge 6-6abc$ isn't necessarily true. For example, consider $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$ . In general, for Olympiads if you get a very large bound (in your solution, $15abc>-11$ is a seriously generous bound), there's a good chance your solution is wrong somewhere.

Edit: Anyways, here's my finish from @above's solution. We have that
$\begin{align*}& 2(a^3+b^3+c^3-3abc)\geq \frac{1-27abc}{2} \\ \iff & 4a^3+4b^3+4c^3\geq 1-15abc \\ \iff & 4a^3+4b^3+4c^3+15abc\geq (a+b+c)^3 \\ \iff & a^3+b^3+c^3+3abc\geq \sum_{cyc}a^2b \end{align*}$
This is just 1st degree Schur's, so we have $2(a^3+b^3+c^3-3abc)\geq \frac{1-27abc}{2}$ which finishes the result from above.

This post has been edited 2 times. Last edited by MortemEtInteritum, Jul 30, 2020, 4:55 AM

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