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#1 h Apr 9, 2025, 5:37 PM • 1 Y

Y by cubres

Find all real numbers $k$ such that the following inequality holds for all $a, b, c \geq 0$ :

$ab + bc + ca \leq \frac{(a + b + c)^2}{3} + k \cdot \max \{ (a - b)^2, (b - c)^2, (c - a)^2 \} \leq a^2 + b^2 + c^2$

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kokcio

#2 h Apr 12, 2025, 7:39 PM • 1 Y

Y by SlovEcience

$\frac{(a+b+c)^2}{3}\geq ab+bc+ca$ , so all non-negative values of $k$ satisfy left inequality. For now, we will now consider only write inequality. WLOG $(a-b)^2=max\{(a-b)^2,(b-c)^2,(c-a)^2\}$ . Then we multiply right inequality to get $3k(a-b)^2\leq (a-b)^2+(b-c)^2+(c-a)^2$ , but $(b-c)^2+(c-a)^2\geq\frac{(a-b)^2}{2}$ with equality for $c=\frac{|b-a|}{2}$ . Therefore, we have that $3k\leq\frac{3}{2}$ , so $k\leq\frac{1}{2}$ . Similarly, we can get $k\geq-\frac{1}{4}$ , so $k\in[-\frac{1}{4},\frac{1}{2}]$

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#3 h Apr 13, 2025, 2:42 PM

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I think k=0 is an answer too

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#4 h Apr 13, 2025, 2:42 PM

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kokcio wrote:

$\frac{(a+b+c)^2}{3}\geq ab+bc+ca$ , so all non-negative values of $k$ satisfy left inequality. For now, we will now consider only write inequality. WLOG $(a-b)^2=max\{(a-b)^2,(b-c)^2,(c-a)^2\}$ . Then we multiply right inequality to get $3k(a-b)^2\leq (a-b)^2+(b-c)^2+(c-a)^2$ , but $(b-c)^2+(c-a)^2\geq\frac{(a-b)^2}{2}$ with equality for $c=\frac{|b-a|}{2}$ . Therefore, we have that $3k\leq\frac{3}{2}$ , so $k\leq\frac{1}{2}$ . Similarly, we can get $k\geq-\frac{1}{4}$ , so $k\in[-\frac{1}{4},\frac{1}{2}]$

Thank you so much for you solution

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