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sqing   3
N Mar 21, 2025 by sqing
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Let $ a,b,c\geq 2.$ Prove that
$$ (a+1)(b+2)(c +1)-3 abc\leq 12$$$$ (a+1)(b+2)(c +1)-\frac{7}{2}abc\leq 8$$$$ (a+1)(b+3)(c +1)-\frac{15}{4}abc\leq 15$$$$ (a+1)(b+3)(c +1)-4abc\leq 13$$
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sqing
Mar 21, 2025
sqing
Mar 21, 2025
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sqing
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sqing
#1 h  Mar 21, 2025, 4:17 AM
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Let $ a,b,c\geq 2.$ Prove that
$$ (a+1)(b+2)(c +1)-3 abc\leq 12$$$$ (a+1)(b+2)(c +1)-\frac{7}{2}abc\leq 8$$$$ (a+1)(b+3)(c +1)-\frac{15}{4}abc\leq 15$$$$ (a+1)(b+3)(c +1)-4abc\leq 13$$
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sqing
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sqing
#2 h  Mar 21, 2025, 4:26 AM
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Let $ a,b,c\geq 2.$ Prove that
$$(a+2)(b+1)(c +2)-4a b c \leq 16$$$$ (a+2)(b+3)(c +2)-5a b c \leq 40$$$$ (a+3)(b+1)(c +3)-\frac{25}{4}a b c \leq 25$$$$ (a+3)(b+2)(c +3)-\frac{25}{4}a b c \leq 50$$
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lbh_qys
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lbh_qys
#3 h  Mar 21, 2025, 6:07 AM
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sqing wrote:
Let $ a,b,c\geq 2.$ Prove that
$$ (a+1)(b+2)(c +1)-3 abc\leq 12$$

Let
\[ f(a,b,c) = (a+1)(b+2)(c+1) - 3abc, \]so that
\[ f_a(a,b,c) = (b+2)(c+1) - 3bc \leq 2b\cdot\frac{3}{2}c - 3bc = 0. \]This indicates that \(f\) is monotonically decreasing with respect to \(a\). By the same reasoning, \(f\) is monotonically decreasing with respect to both \(b\) and \(c\). Consequently,
\[ f(a,b,c) \leq f(2,2,2) = 12. \]
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sqing
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#4 h  Mar 21, 2025, 7:05 AM
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Very nice.Thank bh_qys.
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