inequality

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

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#1 h Mar 29, 2025, 5:50 AM

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Let a,b,c\ge 0: ab+bc+ca>0 then prove \frac{3a-b-c}{b^2+c^2}+\frac{3b-c-a}{c^2+a^2}+\frac{3c-a-b}{a^2+b^2}\ge \frac{3}{2}\cdot\frac{a+b+c}{ab+bc+ca}

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#2 h Mar 29, 2025, 5:53 AM

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JK1603JK wrote:

Let $a,b,c\ge 0$ : $ab+bc+ca>0$ then prove $\frac{3a-b-c}{b^2+c^2}+\frac{3b-c-a}{c^2+a^2}+\frac{3c-a-b}{a^2+b^2}\ge \frac{3}{2}\cdot\frac{a+b+c}{ab+bc+ca}$

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#3 h Mar 31, 2025, 7:11 AM

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Thank you for converting it to latex

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#4 h Mar 31, 2025, 8:03 AM

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The original inequality is equivalent to

$\sum \frac{a-b-c}{b^2+c^2} + 2 \sum \frac{a}{b^2+c^2} \geq \frac{3(a+b+c)}{2(ab+bc+ca)}.$
By the Cauchy–Schwarz inequality, we have

$\sum \frac{a}{b^2+c^2} \geq \frac{(\sum a)^2}{\sum a(b^2+c^2)}.$
Also, notice that

$\sum a(b^2+c^2) = a(b^2+c^2) + (a^2+bc)(b+c).$
Hence,

$\left(\sum a(b^2+c^2)\right)\left(\sum \frac{a-b-c}{b^2+c^2}\right) = \sum (a-b-c) \frac{a(b^2+c^2) + (a^2+bc)(b+c)}{b^2+c^2}$ $=\sum (a-b-c)a + \sum \frac{(a-b-c)(a^2+bc)(b+c)}{b^2+c^2}.$
Moreover, since

$(a-b-c)(a^2+bc) = a(a-b)(a-c) - bc(b+c),$
we have

$\sum (a-b-c)a + \sum \frac{(a-b-c)(a^2+bc)(b+c)}{b^2+c^2} = \sum a^2 - 2\sum ab + \sum \frac{a(a-b)(a-c)(b+c)}{b^2+c^2} - \sum \frac{bc(b+c)^2}{b^2+c^2}.$
Here, the term

$\sum \frac{a(a-b)(a-c)(b+c)}{b^2+c^2}$
is a Schur-like inequality. Assuming $a \geq b \geq c$ , we have

$\frac{a(b+c)}{b^2+c^2} \geq \frac{b(a+c)}{a^2+c^2},$
which suffices to show that it is nonnegative.

Furthermore,

$\sum \frac{bc(b+c)^2}{b^2+c^2} \leq \sum bc\cdot\frac{2(b^2+c^2)}{b^2+c^2} = 2\sum bc.$
Thus,

$\sum a^2 - 2\sum ab + \sum \frac{a(a-b)(a-c)(b+c)}{b^2+c^2} - \sum \frac{bc(b+c)^2}{b^2+c^2} \geq \sum a^2 - 2\sum ab + 0 - 2\sum bc = \sum a^2 - 4\sum ab.$
Therefore,

$\sum \frac{a-b-c}{b^2+c^2} \geq \frac{\sum a^2 - 4\sum ab}{\sum a(b^2+c^2)}.$
It remains to prove that

$\frac{\sum a^2 - 4\sum ab + 2(\sum a)^2 }{\sum a(b^2+c^2)} \geq \frac{3(a+b+c)}{2(ab+bc+ca)}.$
This simplifies to

$\sum ab(a-b)^2 \geq 0,$
which is obviously true. Hence, the original inequality holds.

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