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Inequalities
sqing   16
N Apr 29, 2025 by martianrunner
Let $ a,b \in [0 ,1] . $ Prove that
$$\frac{a}{ 1-ab+b }+\frac{b }{ 1-ab+a } \leq 2$$$$ \frac{a}{ 1+ab+b^2 }+\frac{b }{ 1+ab+a^2 }+\frac{ab }{2+ab }  \leq 1$$$$\frac{a}{ 1-ab+b^2 }+\frac{b }{ 1-ab+a^2 }+\frac{1 }{1+ab  }\leq \frac{5}{2}$$$$\frac{a}{ 1-ab+b^2 }+\frac{b }{ 1-ab+a^2 }+\frac{1 }{1+2ab  }\leq \frac{7}{3}$$$$\frac{a}{ 1+ab+b^2 }+\frac{b }{ 1+ab+a^2 } +\frac{ab }{1+ab }\leq \frac{7}{6 }$$
16 replies
sqing
Apr 25, 2025
martianrunner
Apr 29, 2025
Inequalities
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sqing
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#1
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Let $ a,b \in [0 ,1] . $ Prove that
$$\frac{a}{ 1-ab+b }+\frac{b }{ 1-ab+a } \leq 2$$$$ \frac{a}{ 1+ab+b^2 }+\frac{b }{ 1+ab+a^2 }+\frac{ab }{2+ab }  \leq 1$$$$\frac{a}{ 1-ab+b^2 }+\frac{b }{ 1-ab+a^2 }+\frac{1 }{1+ab  }\leq \frac{5}{2}$$$$\frac{a}{ 1-ab+b^2 }+\frac{b }{ 1-ab+a^2 }+\frac{1 }{1+2ab  }\leq \frac{7}{3}$$$$\frac{a}{ 1+ab+b^2 }+\frac{b }{ 1+ab+a^2 } +\frac{ab }{1+ab }\leq \frac{7}{6 }$$
This post has been edited 4 times. Last edited by sqing, Apr 25, 2025, 9:53 AM
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jestrada
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#2
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$$\frac{a}{ 1-ab+b }+\frac{b }{ 1-ab+a } \leq 2$$
Let $c=1-ab$, so we have to prove $\frac{a}{b+c}+\frac{b}{a+c}\leq 2$, which leads to $a(a+c)+b(b+c)\leq 2(a+c)(b+c)$, or after expanding and simplifying:
$a^2+b^2\leq 2ab+ac+bc+c^2$. Note that $c^2+2ab=1+a^2b^2$, therefore we have to prove $a^2+b^2\leq 1+a^2b^2 +(a+b)c$, or
$(a^2-1)(b^2-1)+(a+b)c\geq0$, which is true since $a,b\in[0,1]$ and $c\geq0$. Equality holds iff $a=b=1$.
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sqing
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#3
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Very good.Thanks.
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DAVROS
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sqing wrote:
Let $ a,b \in [0 ,1] . $ Prove that $ \frac{a}{ 1+ab+b^2 }+\frac{b }{ 1+ab+a^2 }+\frac{ab }{2+ab }  \leq 1$
solution
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sqing
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#5
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Very very nice.Thank DAVROS.
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sqing
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#6
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Let $ a,b,c,d\geq -1 $ and $ a+b+c+d=2. $ Prove that
$$ a + bc + cd + bcd \leq 6$$$$a -ab+bc-cd+ d\leq 9$$$$  a - bc + cd - bcd \leq \frac{73}{8}$$$$a^2+2ab- bc+ 2cd+ d^2\leq\frac{77}{4}$$$$ a^2+2ab-2bc+2cd+ d^2 \leq 20$$$$a^2+2ab-2bc+ cd+ d^2\leq 21$$
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DAVROS
1718 posts
#7
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sqing wrote:
Let $ a,b,c,d\geq -1 $ and $ a+b+c+d=2. $ Prove that $ a + bc + cd + bcd \leq 6$
solution
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SineWaveSage2009
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#8
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these were fun to solve ngl
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sqing
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#9
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Very very nice.Thank DAVROS.
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sqing
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#10
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Let $ a,b\geq 0 $ and $ a+b=2. $ Prove that
$$ a^3b^2( 3 a+ b ) \leq \frac{107+51\sqrt{17}}{64}$$$$ a^3b^2( 2a+b) \leq \frac{2117\sqrt{73}-12881}{1458}$$$$a^2b( 3a+ b ) \leq \frac{827+73\sqrt{73}}{256}$$$$a^2b( \frac 32a+ b ) \leq \frac{827-73\sqrt{73}}{64}$$Let $ a,b\geq 0 $ and $ a+b+ab=3. $ Prove that
$$a^2b ( 2a +  b ) \leq 46+13\sqrt{13}-\frac{1}{9} \sqrt{307014+85254\sqrt{13}}$$
This post has been edited 2 times. Last edited by sqing, Apr 27, 2025, 3:02 AM
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sqing
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#11
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Let $ a,b\geq 0 $ and $ a^2b( 2a+ b ) \geq 4 . $ Prove that
$$a+b\geq 2 $$Let $ a,b\geq 0 $ and $ a^2b^2(a^3+ab+b^3) \geq 3 . $ Prove that
$$a+b\geq 2 $$Let $ a,b\geq 0 $ and $ a^2b^2( 2a^2+b^2)(a^2+2b^2) \geq 9 . $ Prove that
$$a+b\geq 2 $$Let $ a,b\geq 0 $ and $ ab^2( b +1) \geq 4 . $ Prove that
$$a+b+ab\geq 3$$Let $ a,b\geq 0 $ and $ ab( b +1) \geq \frac{9}{4} . $ Prove that
$$a+b+ab\geq 3$$Let $ a,b\geq 0 $ and $a^2b ( a+b^2 ) \geq \frac{76}{27} . $ Prove that
$$a+b+ab\geq 3$$
This post has been edited 1 time. Last edited by sqing, Apr 27, 2025, 12:51 PM
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DAVROS
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#12
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sqing wrote:
Let $ a,b,c,d\geq -1 $ and $ a+b+c+d=2. $ Prove that $a^2+2ab- bc+ 2cd+ d^2\leq\frac{77}{4}$
solution
This post has been edited 1 time. Last edited by DAVROS, Apr 29, 2025, 1:32 PM
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sqing
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#13
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Very very nice.Thank DAVROS.
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DAVROS
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#14
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sqing wrote:
Let $ a,b\geq 0 $ and $ a+b=2. $ Prove that $ a^3b^2( 2a+b) \leq \frac{2117\sqrt{73}-12881}{1458}$
solution
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sqing
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#15
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Very very nice.Thank DAVROS.
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ReticulatedPython
716 posts
#16
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Hi sqing, slightly off topic but where do you get the inequalities from? Do you create them by yourself or do you source them from math competitions?
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martianrunner
212 posts
#17
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I think he just spends a few hours creating inequalities with the givens lol

although I could be wrong, I think this is the most reasonable possibility since the amount of inequalities that he's posted far exceeds the ones that people have solved and also he posts like five equations per given bound
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