A hard inequality

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A hard inequality

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#1 h Mar 23, 2025, 1:40 AM

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Let a,b,c\ge 0: a+b+c=3. Prove \frac{1}{abc}+\frac{12}{a^2b+b^2c+c^2a}\ge 5.

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#2 h Mar 23, 2025, 2:21 AM

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JK1603JK wrote:

Let $a,b,c\ge 0$ : $a+b+c=3$ . Prove $\frac{1}{abc}+\frac{12}{a^2b+b^2c+c^2a}\ge 5.$

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sqing

#3 h Mar 23, 2025, 2:25 AM

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JK1603JK wrote:

Let $a,b,c\ge 0: a+b+c=3.$ Prove $\frac{1}{abc}+\frac{12}{a^2b+b^2c+c^2a}\ge 5.$

If $a,b,c$ are positive real numbers such that $a + b + c = 3$ , then
$\frac{24}{a^2b+b^2c+c^2a} +\frac{1}{abc} \ge 9$ https://artofproblemsolving.com/community/c6h299899p5941489
https://artofproblemsolving.com/community/c6h1300970p6930092

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