An inequality with two equality cases

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High School Math Grades 9-12, Ages 13-18

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An inequality with two equality cases

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#1 h Apr 1, 2025, 8:14 AM

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Let a,b,c\ge 0: a+b+c>0 then prove \frac{b+c-2a}{2a+b+c}+\frac{c+a-2b}{2b+c+a}+\frac{a+b-2c}{2c+a+b}\ge \frac{4(a^2+b^2+c^2-ab-bc-ca)}{5(a^{2}+b^{2}+c^{2})+6(ab+bc+ca)}

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#2 h Apr 1, 2025, 10:23 AM

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JK1603JK wrote:

Let $a,b,c\ge 0: a+b+c>0$ then prove $\frac{b+c-2a}{2a+b+c}+\frac{c+a-2b}{2b+c+a}+\frac{a+b-2c}{2c+a+b}\ge \frac{4(a^2+b^2+c^2-ab-bc-ca)}{5(a^{2}+b^{2}+c^{2})+6(ab+bc+ca)}$

Since
$\sum \frac{b+c-2a}{b+c+2a} = \sum \frac{b-a}{b+c+2a} + \sum \frac{a-b}{a+c+2b} = \sum \frac{(a-b)^2}{(b+c+2a)(a+c+2b)},$ it follows that
$\begin{aligned} \left(5\sum a^2 + 6\sum ab\right) \cdot \sum \frac{(a-b)^2}{(b+c+2a)(a+c+2b)} - 4\left(\sum a^2 - \sum ab\right) &=\sum (a-b)^2\left(\frac{5\sum a^2 + 6\sum ab}{(b+c+2a)(a+c+2b)} - 2\right)\\[1mm] &=\sum \frac{(a-b)^2\left(a^2 + b^2 - 4ab + 3c^2\right)}{(b+c+2a)(a+c+2b)}. \end{aligned}$
Noting that
$a^2 + b^2 - 4ab + 3c^2 = (a+b-3c)^2 + 6(b-c)(c-a) \geq 6(b-c)(c-a),$ we have
$\sum \frac{(a-b)^2\left(a^2 + b^2 - 4ab + 3c^2\right)}{(b+c+2a)(a+c+2b)} \geq 6\sum \frac{(a-b)^2(b-c)(c-a)}{(b+c+2a)(a+c+2b)}.$
Moreover,
$6\sum \frac{(a-b)^2(b-c)(c-a)}{(b+c+2a)(a+c+2b)} =\frac{6(a-b)(b-c)(c-a)}{(b+c+2a)(a+c+2b)(a+b+2c)} \cdot \sum (a-b)(a+b+2c)=0.$
So the original inequality holds.

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