MM 2201 (Symmetric Inequality with Weird Sharp Case)

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MM 2201 (Symmetric Inequality with Weird Sharp Case)

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Source: Mathematics Magazine Volume 97 (2024), Issue 4: https://doi.org/10.1080/0025570X.2024.2393998

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#1 h Apr 10, 2025, 6:19 PM

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2201. Proposed by Leonard Giugiuc, Drobeta-Turnu Severin, Romania. Find all real numbers $K$ such that
$a^2 + b^2 + c^2 - 3 \geq K(a + b + c - 3)$ for all nonnegative real numbers $a$ , $b$ , and $c$ with $abc \leq 1$ .

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#2 h Apr 10, 2025, 7:11 PM

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Taking $(e,1,\frac{1}{e})$ we get $e^4-2e^2+1\geq e(e^2-2e+1)K$ .
So $K \leq e-2+ \frac{1}{e}$ .
Since we can choose $e$ infinitely close to 1, we have $K \leq 2$ .
Taking $(1-e,1-e,1-e)$ we get $3e^2-6e \geq -3eK$
So $K \geq 2-e$ .
Since we can choose $e$ infinitely small, we have $K \geq 2$ .
If $K = 2$ we have $a^2+b^2+c^2-3 \geq (2a-1)+(2b-1)+(2c-1)-3 = 2(a+b+c-3)$ .
Answer: $K=2$ .

This post has been edited 2 times. Last edited by CHESSR1DER, Apr 10, 2025, 7:46 PM

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