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 31, 2025, 7:11 AM

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Let a,b,c\ge 0: ab+bc+ca>0 then prove \frac{4ab+5c^2}{a+b}+\frac{4bc+5a^2}{b+c}+\frac{4ca+5b^2}{c+a}\ge \frac{3}{2}\cdot\frac{(a+b+c)^3}{ab+bc+ca}.

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#2 h Mar 31, 2025, 9:27 AM

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JK1603JK wrote:

Let $a,b,c\ge 0$ : $ab+bc+ca>0$ then prove $\frac{4ab+5c^2}{a+b}+\frac{4bc+5a^2}{b+c}+\frac{4ca+5b^2}{c+a}\ge \frac{3}{2}\cdot\frac{(a+b+c)^3}{ab+bc+ca}.$

It appears that this inequality can be proven using the following generalization of the Iran 96 inequality

$\frac{15}{16} \cdot \frac{\sum a(a-b)(a-c)}{\prod (a+b)} \leq \left(\sum ab\right)\left(\sum \frac{1}{(a+b)^2}\right) -\frac 94\leq \frac{\sum a(a-b)(a-c)}{\prod (a+b)}$

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

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Can you explain more detail?

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#4 h Apr 1, 2025, 9:22 AM

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We have
$\frac{4ab+5c^2}{a+b}=\frac{4(a+c)(b+c)}{a+b}+\frac{c^2}{a+b}-4c.$ According to the above inequality, we have
$4\Bigl(\sum ab\Bigr) \sum \frac{(a+c)(b+c)}{a+b}=4\Bigl(\sum ab\Bigr)\prod (a+b)\sum\frac{1}{(a+b)^2}\ge 9\prod (a+b)+\frac{15}{4}\sum a(a-b)(a-c)\ge 9\prod (a+b)+\frac{1}{2}\sum a(a-b)(a-c),$ and
$\Bigl(\sum ab\Bigr)\sum\frac{c^2}{a+b}=\sum c^3+abc\sum\frac{c}{a+b}\ge \sum c^3+\frac{3}{2}abc.$ Thus, it suffices to prove
$9\prod (a+b)+\frac{1}{2}\sum a(a-b)(a-c)+\sum c^3+\frac{3}{2}abc\ge \frac{3}{2}(a+b+c)^3+4(ab+bc+ca)(a+b+c).$ In fact, this is an identity; hence, it is only necessary to prove the generalized form of the Iran 96 inequality as stated above.

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