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#1 h Mar 23, 2025, 12:32 PM

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This post has been edited 1 time. Last edited by Cobedangiu, Mar 23, 2025, 12:32 PM

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#2 h Mar 23, 2025, 12:41 PM

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Let $t = bc$ , then $8 = a^2 + b^2 + c^2 \geq 2bc$ so $t = bc \leq 4$ .

Using the Cauchy-Schwarz inequality, we get:

$[4(a+b+c)-abc]^2 = [(4-bc)a+4(b+c)]^2 \leq [(4-bc)^2+4^2][a^2+(b+c)^2]$
$= (t^2-8t+32)(2t+8) = 2t^3-8t^2+256 = 2t^2(t-4)+256 \leq 256.$
Therefore $4(a+b+c)-abc \leq 16$ and $4(a+b+c-4) \leq abc$ .

We have equality iff $a=0, b=c=2$ or permutations.

This post has been edited 1 time. Last edited by pooh123, Mar 23, 2025, 1:24 PM
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