A non-homogenous inequality

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A non-homogenous inequality

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#1 h Mar 30, 2025, 9:20 AM

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Let a,b,c>0 with abc=k^3 and k>0 then prove \frac{a^4}{k^9+pa^3}+\frac{b^4}{k^9+pb^3}+\frac{c^4}{k^9+pc^3}\ge \frac{3k}{k^6+p}.

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#2 h Mar 30, 2025, 12:06 PM

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JK1603JK wrote:

Let $a,b,c>0$ with $abc=k^3$ and $k>0$ then prove $\frac{a^4}{k^9+pa^3}+\frac{b^4}{k^9+pb^3}+\frac{c^4}{k^9+pc^3}\ge \frac{3k}{k^6+p}.$

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#3 h Mar 31, 2025, 5:09 AM • 1 Y

Y by MihaiT

Let
$a, b, c = \frac{kx}{y},\ \frac{ky}{z},\ \frac{kz}{x},$ then the original inequality is equivalent to
$\sum \frac{x^4}{k^6 y^4 + p x^3 y} \geq \frac{3}{k^6 + p}.$
By the Cauchy-Schwarz inequality,
$\sum \frac{x^4}{k^6 y^4 + p x^3 y} \geq \frac{\left(\sum \frac{x^2}{y}\right)^2}{k^6 \sum y^2 + p \sum \frac{x^3}{y}},$ so it suffices to prove
$\left(\sum \frac{x^2}{y}\right)^2 \geq 3 \sum \frac{x^3}{y} \geq 3 \sum y^2.$
The proof of the last inequality can be found at https://artofproblemsolving.com/community/c6h153560p864806.

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