Interesting inequality

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sqing

#1 h Mar 20, 2025, 4:34 AM

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Let $a,b\geq 0 .$ Prove that
$a^4+b^4 +kab\geq\left(\sqrt{k(k+2)}-k\right)ab(a+b+k)$ Where $k>0 .$
$a^4+b^4 +ab\geq (\sqrt 3-1)ab(a+b+1)$ $a^4+b^4 +2ab\geq 2(\sqrt 2-1)ab(a+b+2)$

This post has been edited 1 time. Last edited by sqing, Mar 20, 2025, 7:39 AM

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#2 h Mar 20, 2025, 7:10 AM

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sqing wrote:

Let $a,b\geq 0 .$ Prove that
$a^4+b^4 +kab\geq\left(\sqrt{k(k+2)}-k\right)ab(a+b+k)$ Where $k>0 .$
$a^4+b^4 +ab\geq (\sqrt 3-1)ab(a+b+1)$ $a^4+b^4 +2ab\geq 2(\sqrt 2-1)ab(a+b+2)$

By the AM-GM inequality,
$a^4+b^4+kab\ge\frac{(a+b)^2ab}{2}+kab=\frac{\Bigl[(a+b+k)^2+k(k+2)-2k(a+b+k)\Bigr]ab}{2}\ge\frac{\Bigl[2\sqrt{k(k+2)}(a+b+k)-2k(a+b+k)\Bigr]ab}{2}=(\sqrt{k(k+2)}-k)ab(a+b+k).$

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sqing

#3 h Mar 20, 2025, 7:37 AM

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very nice.Thank lbh_qys.Sorry.

This post has been edited 1 time. Last edited by sqing, Mar 20, 2025, 7:40 AM

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