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sqing   3
N 2 hours ago by sqing
Let $ a,b> 0$ and $ a+b=1 . $ Prove that
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4096}}{1+ ka^7b^7}$$Where $\frac{8192}{3}\geq k>0 .$
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{\frac{14}{3}}{1+ \frac{8192}{3}a^7b^7}$$
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sqing
3 hours ago
sqing
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sqing   0
3 hours ago
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
3 hours ago
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sqing   6
N Mar 17, 2025 by sqing
Let $ a, b\geq 0 $ and $ \frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1} \geq 3 $. Prove that
$$ a+b+a^2+b^2 \geq 28-6\sqrt{21}$$Let $ a, b\geq 0 $ and $ \frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1} \geq 4 $. Prove that
$$ a+b+a^2+b^2 \geq 10-4\sqrt 5$$Let $ a, b\geq 0 $ and $ \frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1} \geq 5 $. Prove that
$$ a+b+a^2+b^2 \geq \frac {2(26-5\sqrt{13})}{9} $$Let $ a, b\geq 0 $ and $\frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1}\geq 3+ \sqrt {5} $. Prove that
$$ a+b+a^2+b^2 \geq 2$$
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sqing
Mar 14, 2025
sqing
Mar 17, 2025
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sqing
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sqing
#1 h  Mar 14, 2025, 8:09 AM
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Let $ a, b\geq 0 $ and $ \frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1} \geq 3 $. Prove that
$$ a+b+a^2+b^2 \geq 28-6\sqrt{21}$$Let $ a, b\geq 0 $ and $ \frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1} \geq 4 $. Prove that
$$ a+b+a^2+b^2 \geq 10-4\sqrt 5$$Let $ a, b\geq 0 $ and $ \frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1} \geq 5 $. Prove that
$$ a+b+a^2+b^2 \geq \frac {2(26-5\sqrt{13})}{9} $$Let $ a, b\geq 0 $ and $\frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1}\geq 3+ \sqrt {5} $. Prove that
$$ a+b+a^2+b^2 \geq 2$$
This post has been edited 2 times. Last edited by sqing, Mar 14, 2025, 8:31 AM
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#2 h  Mar 14, 2025, 8:48 AM
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Let $ a, b\geq 0 $ and $\frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1}\geq 4 $. Prove that
$$ a+b+a^2+ab+b^2 \geq \frac{27-11\sqrt {5}}{2} $$Let $ a, b\geq 0 $ and $\frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1}\geq \frac{2}{5}(7+ 2\sqrt {7}) $. Prove that
$$ a+b+a^2+ab+b^2 \geq 2$$
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lbh_qys
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#3 h  Mar 14, 2025, 11:02 AM
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sqing wrote:
Let $ a, b\geq 0 $ and $ \frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1} \geq 3 $. Prove that
$$ a+b+a^2+b^2 \geq 28-6\sqrt{21}$$
WLOG \( a \geq b \).

If \( a+b \geq 1 \), then
\[ a+b+a^2+b^2 \geq 1 \geq 28-6\sqrt{21}. \]
If \( a+b < 1 \), then
\[ \frac{1}{b^2-b+1} \leq \frac{1}{a^2-a+1}, a^2 + a + 1 \geq b^2 + b + 1 \]which implies that
\[ \frac{a^2+a+1}{a^2-a+1}+\frac{b^2+b+1}{b^2-b+1} \geq \frac{a^2+a+1}{b^2-b+1}+\frac{b^2+b+1}{a^2-a+1} \geq 3. \]
Thus, by the Cauchy–Schwarz inequality,
\[ 3 \geq 6-\left(\frac{a^2+a+1}{a^2-a+1}+\frac{b^2+b+1}{b^2-b+1}\right) =\frac{2(a-1)^2}{a^2-a+1}+\frac{2(b-1)^2}{b^2-b+1} \geq \frac{2(1-a+1-b)^2}{a^2-a+1+b^2-b+1}. \]
Let \( u = a+b \) and \( v = ab \). Then,
\[ 3\left(u^2-u+1-2v\right) \geq 2\,(2-u)^2. \]
Hence,
\[ u^2-4v \geq \frac{u^2-10u+4}{3}. \]
Since
\[ a+b+a^2+b^2 = u^2-2v+u = \frac{u^2-4v}{2}+\frac{u^2}{2}+u, \]if \( u \leq 5-\sqrt{21} < \frac{1}{2} \), then
\[ \frac{u^2-4v}{2}+\frac{u^2}{2}+u \geq \frac{u^2-10u+4}{6}+\frac{u^2}{2}+u =\frac{2}{3}\,(u^2-u+1). \]
Because \( u^2-u+1 \) is monotonically decreasing on \( u\in \left(0,5-\sqrt{21}\right] \), it follows that
\[ \frac{2}{3}\,(u^2-u+1) \geq \frac{2}{3}\Bigl[(5-\sqrt{21})^2-(5-\sqrt{21})+1\Bigr] = 28-6\sqrt{21}. \]
If \( 5-\sqrt{21} < u < 1 \), then since \( u^2\geq 4v \), we have
\[ \frac{u^2-4v}{2}+\frac{u^2}{2}+u \geq \frac{u^2}{2}+u \geq \frac{(5-\sqrt{21})^2}{2}+(5-\sqrt{21}) = 28-6\sqrt{21}. \]
Therefore, in all cases,
\[ a+b+a^2+b^2 \geq 28-6\sqrt{21}. \]
This post has been edited 5 times. Last edited by lbh_qys, Mar 14, 2025, 11:06 AM
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#4 h  Mar 14, 2025, 1:43 PM
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Very very nice.Thank lbh_qys.
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DAVROS
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#5 h  Mar 16, 2025, 5:52 PM
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sqing wrote:
Let $ a, b\geq 0 $ and $ \frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1} \geq 5 $. Prove that $ a+b+a^2+b^2 \geq \frac {2(26-5\sqrt{13})}{9} $
solution
This post has been edited 2 times. Last edited by DAVROS, Mar 16, 2025, 9:24 PM
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#6 h  Mar 17, 2025, 2:39 AM
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Very very nice.Thank DAVROS.
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#7 h  Mar 17, 2025, 9:22 AM
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Let $ a , b\geq 0 $ and $ \frac{1}{a^2+1}+\frac{1}{b^2+1} \le1-\frac{1 }{\sqrt2} . $ Show that$$ a+b+ab\geq5+4\sqrt 2$$Let $ a , b\geq 0 $ and $ \frac{1}{a^2+1}+\frac{1}{b^2+1} \ge1+\frac{1 }{\sqrt2} . $ Show that$$ a+b+ab\leq1$$Let $ a , b\geq 0 $ and $ \frac{1}{a^2+1}+\frac{1}{b^2+1}\ge \frac{2(5+2\sqrt3) }{13} . $ Show that$$ a+b+ab\leq2$$Let $ a , b\geq 0 $ and $ \frac{1}{a^2+1}+\frac{1}{b^2+1} \le \frac{15+\sqrt 5 }{10}. $ Show that$$ a+b+ab+a^2+b^2\geq 1$$Let $ a , b $ be reals such that $ \frac{1}{a^2+1}+\frac{1}{b^2+1}\ge \frac{4(4+\sqrt3) }{13}. $ Show that$$ a+b+a^2+b^2\leq 1$$Let $ a , b $ be reals such that $ \frac{1}{a^2+1}+\frac{1}{b^2+1}\ge \frac{9 }{5}. $ Show that$$ a+b+ab+a^2+b^2\leq 1$$
This post has been edited 6 times. Last edited by sqing, Mar 17, 2025, 1:07 PM
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