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lgx57   1
N 36 minutes ago by sqing
Let $a,b,c>0$,$\frac{a^2+b^2+c^2}{ab+bc+ca}=2$, find the minimum of

$$\frac{a^3+b^3+c^3}{abc}$$
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sqing   6
N an hour ago by sqing
Let $ a,b,c\geq 0 $ and $a+b+c=1$. Prove that$$a^3b+b^3c+c^3a+\frac{473}{256}abc\le\frac{27}{256}$$Equality holds when $ a=b=c=\frac{1}{3} $ or $ a=0,b=\frac{3}{4},c=\frac{1}{4} $ or $ a=\frac{1}{4} ,b=0,c=\frac{3}{4} $
or $ a=\frac{3}{4} ,b=\frac{1}{4},c=0. $
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sqing   12
N Mar 21, 2025 by sqing
Let $ a,b $ be real numbers such that $ a + b \geq |ab + 1|. $ Prove that$$ a^3 + b^3 \geq |a^3 b^3 + 1|$$Let $ a,b $ be real numbers such that $ 2(a + b ) \geq |ab + 1|. $ Prove that$$26( a^3 + b^3) \geq |a^3 b^3 + 1|$$Let $ a,b $ be real numbers such that $ 4(a + b) \geq 3|ab + 1|. $ Prove that$$148(a^3 + b^3) \geq27 |a^3 b^3 + 1|$$
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sqing
Mar 8, 2025
sqing
Mar 21, 2025
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sqing
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#1 h  Mar 8, 2025, 12:43 PM
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Let $ a,b $ be real numbers such that $ a + b \geq |ab + 1|. $ Prove that$$ a^3 + b^3 \geq |a^3 b^3 + 1|$$Let $ a,b $ be real numbers such that $ 2(a + b ) \geq |ab + 1|. $ Prove that$$26( a^3 + b^3) \geq |a^3 b^3 + 1|$$Let $ a,b $ be real numbers such that $ 4(a + b) \geq 3|ab + 1|. $ Prove that$$148(a^3 + b^3) \geq27 |a^3 b^3 + 1|$$
This post has been edited 1 time. Last edited by sqing, Mar 8, 2025, 12:54 PM
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pennypc123456789
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#2 h  Mar 8, 2025, 1:02 PM
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sqing wrote:
Let $ a,b $ be real numbers such that $ a + b \geq |ab + 1|. $ Prove that$$ a^3 + b^3 \geq |a^3 b^3 + 1|$$
Let $ a,b $ be real numbers such that $ 4(a + b) \geq 3|ab + 1|. $ Prove that$$148(a^3 + b^3) \geq27 |a^3 b^3 + 1|$$

since $a^2b^2 - ab + 1 > 0 $ , we have :
$|a^3b^3 + 1| = |ab + 1|( a^2b^2 - ab + 1) = |ab + 1| [(ab+1)^2-3ab] \le (a+b)[(a+b)^2-3ab] = (a + b )(a^2-ab+b^2) = a^3 +b^3 . $
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#3 h  Mar 8, 2025, 1:23 PM • 1 Y
Y by pennypc123456789
Good.Thanks.
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DAVROS
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#4 h  Mar 9, 2025, 10:20 AM
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sqing wrote:
Let $ a,b $ be real numbers such that $ 4(a + b) \geq 3|ab + 1|. $ Prove that $148(a^3 + b^3) \geq27 |a^3 b^3 + 1|$
solution
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#5 h  Mar 9, 2025, 2:37 PM
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Very very nice.Thank DAVROS.
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#6 h  Mar 11, 2025, 9:10 AM
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Let $a,b,c \in [0,1].$ Prove that$$-\frac{3}{4}\leq (a-b)(b-2c)(c-3a) \leq 4$$$$-6\leq (a-2b)(b-3c)(c-4a) \leq 9$$$$-4\leq (a- 2b)(b+c)(c-2a) \leq \frac{9}{4}$$
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sqing
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#7 h  Mar 19, 2025, 4:16 AM
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Let $ a,b\ge 0 $ and $ 2a + 2b \ge 4 + ab. $ Prove that
$$a^2 + b^2 +ab \ge 4 $$$$ a^2 + b^2 +ab +a + b \ge 6$$$$ a^2 + b^2 +ab + 2a + 2b \ge 8$$Let $ a,b\ge 1 $ and $ 2a + 2b\ge 4 + ab. $ Prove that
$$a^2 + b^2 +ab \ge 7$$$$ a^2 + b^2 +ab +a + b \ge 10$$$$ a^2 + b^2 +ab + 2ab + 2b \ge 13$$
This post has been edited 1 time. Last edited by sqing, Mar 19, 2025, 4:21 AM
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DAVROS
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#8 h  Mar 21, 2025, 6:57 AM
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sqing wrote:
Let $ a,b\ge 0 $ and $ 2a + 2b \ge 4 + ab. $ Prove that $a^2 + b^2 +ab \ge 4 $
solution
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DAVROS
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#9 h  Mar 21, 2025, 7:31 AM
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sqing wrote:
Let $ a,b\ge 1 $ and $ 2a + 2b\ge 4 + ab. $ Prove that $a^2 + b^2 +ab \ge 7$
solution
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#10 h  Mar 21, 2025, 7:57 AM
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Very very nice.Thank DAVROS.
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sqing
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#11 h  Mar 21, 2025, 8:49 AM
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Let $ a,b,c\geq 1. $ Prove that
$$ (a+1)(b+2)(c +1)-6 abc\leq 6$$$$ (a+1)(b+3)(c +1)-8abc\leq 8$$$$(a+2)(b+1)(c +2)-9a b c \leq 9$$$$ (a+2)(b+3)(c +2)-12a b c \leq 24$$$$ (a+3)(b+1)(c +3)-16a b c \leq 16$$$$ (a+3)(b+2)(c +3)-16a b c \leq 32$$
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DAVROS
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#12 h  Mar 21, 2025, 9:34 AM
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sqing wrote:
Let $ a,b,c\geq 1. $ Prove that $ (a+1)(b+2)(c +1)-6 abc\leq 6$
solution
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#13 h  Mar 21, 2025, 1:12 PM
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Very very nice.Thank DAVROS.
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