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sqing   29
N 41 minutes ago by SomeonecoolLovesMaths
Let $ a,b>0 $ and $ \frac{1}{a}+\frac{1}{b}=1. $ Prove that
$$(a^2-a+1)(b^2-b+1) \geq 9$$$$ (a^2-a+b+1)(b^2-b+a+1) \geq 25$$Let $ a,b>0 $ and $ \frac{1}{a}+\frac{1}{b}=\frac{2}{3}. $ Prove that
$$(a+8)(a^2-a+b+2)(b^2-b+5)\geq1331$$$$(a+10)(a^2-a+b+4)(b^2-b+7)\geq2197$$
29 replies
sqing
Mar 10, 2025
SomeonecoolLovesMaths
41 minutes ago
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sqing   12
N an hour ago 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$$148(a^3 + b^3) \geq27 |a^3 b^3 + 1|$$
12 replies
sqing
Mar 8, 2025
sqing
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sqing   4
N Today at 2:57 AM by sqing
Let $ a,b,c\geq 2 . $ Prove that
$$(a-1)(b^2-2)(c-1) - \frac{1}{2}abc\geq -2$$$$(a^2-2)(b-1)(c^2-2) - \frac{3}{2}abc\geq -6$$
4 replies
sqing
Mar 19, 2025
sqing
Today at 2:57 AM
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sqing
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sqing
#1 h  Mar 19, 2025, 1:42 PM
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Let $ a,b,c\geq 2 . $ Prove that
$$(a-1)(b^2-2)(c-1) - \frac{1}{2}abc\geq -2$$$$(a^2-2)(b-1)(c^2-2) - \frac{3}{2}abc\geq -6$$
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no_room_for_error
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#2 h  Wednesday at 6:07 PM
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sqing wrote:
Let $ a,b,c\geq 2 . $ Prove that
$$(a-1)(b^2-2)(c-1) - \frac{1}{2}abc\geq -2$$

$$ \begin{aligned} (a-1)(b^2-2)(c-1) &\geq 2(a-1)(b-1)(c-1)\\ &=\frac{1}{2}abc-2+\frac{3}{2}(a-2)(b-2)(c-2)+(a-2)(b-2)+(b-2)(c-2)+(c-2)(a-2)\\ &\geq \frac{1}{2}abc-2\\ \end{aligned} $$
Equality occurs when $b=2$ and $(c-2)(a-2)=0$.
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sqing
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#3 h  Yesterday at 2:19 AM
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Very very nice.Thank no_room_for_error.
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sqing
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#4 h  Yesterday at 1:58 PM
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Let $ a,b\geq 0 $ and $ a^2+ab+b^2+a+b=5. $ Prove that
$$ \frac{ (a+b)(ab+1)}{a+b-1} \leq 4$$$$ \frac{ (a+b)(ab+a+b)}{a+b-1} \leq 6$$$$ \frac{a^2b+b^2+a+1}{a+b } \leq \frac{3\sqrt{21}-2}{5}$$$$ \frac{a^2b+b^2+a }{a+b+1} \leq \frac{3\sqrt{21}-8}{5}$$
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sqing
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#5 h  Today at 2:57 AM
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Let $ a,b,c\geq \frac{4}{5}$ and $ a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+8 . $ Prove that
$$ ab+bc +ca\leq 27 $$
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