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Inequalities
sqing   41
N 41 minutes ago by ytChen
Let $a,b>0$ and $ab^2(a+b)=9.$ Prove that
$$2a+5b\geq 2\sqrt[4]{27(3+8\sqrt{6})} $$$$2a+9b\geq 6\sqrt[4]{48\sqrt{2}-39} $$
41 replies
sqing
Oct 31, 2024
ytChen
41 minutes ago
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Inequalities
sqing   5
N Yesterday at 7:55 AM by sqing
Let $ a,b \geq 0 , a^2+b^2= 3 $ and $ a^4+b^4 = 7.$ Prove that $$\frac{15\sqrt 5-11}{2}\leq  a^5+2b^5\leq \frac{15\sqrt 5+11}{2}$$$$  15\sqrt 5-22 \leq a^5 +5b^5\leq 15\sqrt 5+22 $$
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sqing
Nov 14, 2024
sqing
Yesterday at 7:55 AM
Inequalities
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sqing
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Let $ a,b \geq 0 , a^2+b^2= 3 $ and $ a^4+b^4 = 7.$ Prove that $$\frac{15\sqrt 5-11}{2}\leq  a^5+2b^5\leq \frac{15\sqrt 5+11}{2}$$$$  15\sqrt 5-22 \leq a^5 +5b^5\leq 15\sqrt 5+22 $$
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sqing
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#2
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Let $ a ,b>0 $ and $ (a+1)(2b+3)=4 $. Prove that $$2ab+\frac{1}{ab}\geq 21+4\sqrt{3}$$Let $ a ,b>0 $ and $ (a+1)(2b+3)=6 $. Prove that $$2ab+\frac{1}{ab}\geq 11-\frac{14\sqrt{2}}{3}$$
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DAVROS
1470 posts
#3
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sqing wrote:
Let $ a ,b>0 $ and $ (a+1)(2b+3)=6 $. Prove that $2ab+\frac{1}{ab}\geq 11-\frac{14\sqrt{2}}{3}$
solution
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sqing
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#4
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Very very nice.Thank DAVROS.
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sqing
38870 posts
#5
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Let $ a,b,c\in [0,1] $ and $ a+b+c=2 .$ Prove that
$$\dfrac{1}{2}\leq  \dfrac{1-a^2}{a+2}+\dfrac{1-b^2}{b+2}+\dfrac{1-c^2}{c+2} \leq \dfrac{5}{8}$$$$\dfrac{14}{5}\leq\dfrac{2-a^2}{a+1}+\dfrac{2-b^2}{b+1}+\dfrac{2-c^2}{c+1}\leq3$$Let $ a,b,c\in [0,1] $ and $ a+b+c=1 .$ Prove that
$$1\leq\dfrac{1-a^2}{a+2}+\dfrac{1-b^2}{b+2}+\dfrac{1-c^2}{c+2} \leq \dfrac{8}{7}$$$$\dfrac{17}{4}\leq\dfrac{2-a^2}{a+1}+\dfrac{2-b^2}{b+1}+\dfrac{2-c^2}{c+1} \leq \dfrac{9}{2}$$
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sqing
38870 posts
#6
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Let $ a,b,c\in [0,2] $ and $ a+b+c=3 .$ Prove that
$$1+\dfrac{2\sqrt 2}{3}\leq \dfrac{2-a^2}{a+1}+\dfrac{3-b^2}{b+1}+\dfrac{2-c^2}{c+1}\leq \dfrac{17}{6}$$$$ \dfrac{9+4\sqrt 2}{6}\leq \dfrac{3-a^2}{a+1}+\dfrac{2-b^2}{b+1}+\dfrac{3-c^2}{c+1} \leq \dfrac{10}{3}$$
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