An inequality

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

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An inequality

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#1 h Mar 29, 2025, 1:05 PM

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Let a,b,c\ge 0: ab+bc+ca>0 then prove \frac{5ab+c^2}{a+b}+\frac{5bc+a^2}{b+c}+\frac{5ca+b^2}{c+a}\ge 9\cdot\frac{ab+bc+ca}{a+b+c}.

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#2 h Mar 29, 2025, 6:25 PM

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Nooo noo formating hurt my eyes Jek1603k
Let $a,b,c\ge 0: ab+bc+ca>0$ then prove $\frac{5ab+c^2}{a+b}+\frac{5bc+a^2}{b+c}+\frac{5ca+b^2}{c+a}\ge 9\cdot\frac{ab+bc+ca}{a+b+c}.$
this much better. And I pr0bably will not see this again so welcome. If you don't say thank you I will take back you are welcome. I don't want to be that guy who says welcome before other guy says thank you. In Kentucky we are humble.
Your welcome

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#3 h Mar 29, 2025, 6:31 PM

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Nooo noo formating hurt my eyes Jek1603k
Let $a,b,c\ge 0: ab+bc+ca>0$ then prove $\frac{5ab+c^2}{a+b}+\frac{5bc+a^2}{b+c}+\frac{5ca+b^2}{c+a}\ge 9\cdot\frac{ab+bc+ca}{a+b+c}.$
this much better. And I pr0bably will not see this again so welcome. If you don't say thank you I will take back you are welcome. I don't want to be that guy who says welcome before other guy says thank you. In Kentucky we are humble.
Your welcome

They can't latex......

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#4 h Mar 31, 2025, 3:46 AM

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Multiplying both sides of the inequality by $a+b+c$ , we have

$(a+b+c)\frac{a^2 + 5bc}{b+c} = \frac{a(a-b)(a-c)}{b+c} + \frac{4abc}{b+c} + 2a^2 + 5bc,$
so it is equivalent to proving

$\sum \frac{a(a-b)(a-c)}{b+c} + 4abc \sum \frac{1}{b+c} + 2\sum a^2 + 5\sum bc \geq 9 \sum bc.$
According to Schur's inequality,

$\sum \frac{a(a-b)(a-c)}{b+c} = \frac{1}{(a+b)(b+c)(c+a)} \sum a(a^2-b^2)(a^2-c^2) \geq 0,$
so it suffices to prove

$4abc \sum \frac{1}{b+c} \geq 4 \sum bc - 2 \sum a^2.$
This clearly holds by applying the Cauchy–Schwarz inequality,

$\sum \frac{1}{b+c} \geq \frac{9}{\sum (b+c)} = \frac{9}{2(a+b+c)},$
and Schur's inequality

$\frac{9abc}{a+b+c} \geq 2\sum bc - \sum a^2.$
Hence, the original inequality is established.

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