Question

Please present a sharp upper bound or a tight upper bound for $a^{1/q}-b^{1/q}$ ($a\geq b\geq 0$ and $q>1$).

score 119 · Answer 1 · 2018-11-12 11:01:18

119

Views

Please present a sharp upper bound or a tight upper bound for $a^{1/q}-b^{1/q}$ ($a\geq b\geq 0$ and $q>1$).

Published on 12 Nov 2018 - 11:01

score 62 · Answer 2 · 2018-11-12 11:58:57

62

Views

Facing a problem in solving for $x$ from $-2x-1>49$

Published on 12 Nov 2018 - 11:58

score 553 · Answer 3 · 2018-11-12 14:53:06

553

Views

Prove $a < 1/a < b < 1/b$ implies $a < -1$

Published on 12 Nov 2018 - 14:53

score 93 · Answer 4 · 2018-11-12 16:09:13

93

Views

Proof that $\left|\arctan (x)-\frac{π}{4}-\frac{(x-1)}{2}\right| \leq \frac{(x-1)^2}{2}$

Published on 12 Nov 2018 - 16:09

score 72 · Answer 5 · 2026-04-02 19:18:51

72

Views

Solution of $\log_{\frac12}(x) > 4$

Published on 02 Apr 2026 - 19:18

score 103 · Answer 6 · 2026-04-11 16:34:45

103

Views

Solving a system of equations and inequalities

Published on 11 Apr 2026 - 16:34

score 30 · Answer 7 · 2018-11-13 03:48:40

30

Views

a bound on $\|z\|_p$ with high probability for generalized Gaussian vector

Published on 13 Nov 2018 - 3:48

score 98 · Answer 8 · 2018-11-13 07:51:57

98

Views

Inequality that I believe can be conquered with Cauchy Schwarz I.E

Published on 13 Nov 2018 - 7:51

score 61 · Answer 9 · 2026-04-08 17:00:43

61

Views

An inequality with constraints.

Published on 08 Apr 2026 - 17:00

score 196 · Answer 10 · 2018-11-13 11:42:02

196

Views

Sum of fractions Inequality: $ {a\over b} + {b \over c} + {c \over a} \ge 3 $

Published on 13 Nov 2018 - 11:42

score 1.1k · Answer 11 · 2026-04-10 17:13:03

1.1k

Views

The $1997$ IIT JEE problem

Published on 10 Apr 2026 - 17:13

score 353 · Answer 12 · 2018-11-13 15:53:58

353

Views

Proving $(1-a)(1-b)(1-c)(1-d) > 1-a-b-c-d$

Published on 13 Nov 2018 - 15:53

score 53 · Answer 13 · 2026-04-12 12:30:25

53

Views

Inequality in $L^2$

Published on 12 Apr 2026 - 12:30

score 68 · Answer 14 · 2026-03-25 17:42:00

68

Views

Show that $x_{n+1} = \frac{x_n + 1}{n+1}$ is decreasing starting from some $n_0$ and find $n_0$.

Published on 25 Mar 2026 - 17:42

score 58 · Answer 15 · 2026-03-28 15:36:40

58

Views

Proving $\left( 1-\frac{2}{n} \right )^{\frac {n\ln n}{4}}-\left( 1-\frac{1}{n} \right )^{\frac {2n\ln n}{4}}<0$

Published on 28 Mar 2026 - 15:36

List Question

Please present a sharp upper bound or a tight upper bound for $a^{1/q}-b^{1/q}$ ($a\geq b\geq 0$ and $q>1$).

Facing a problem in solving for $x$ from $-2x-1>49$

Prove $a < 1/a < b < 1/b$ implies $a < -1$

Proof that $\left|\arctan (x)-\frac{π}{4}-\frac{(x-1)}{2}\right| \leq \frac{(x-1)^2}{2}$

Solution of $\log_{\frac12}(x) > 4$

Solving a system of equations and inequalities

a bound on $\|z\|_p$ with high probability for generalized Gaussian vector

Inequality that I believe can be conquered with Cauchy Schwarz I.E

An inequality with constraints.

Sum of fractions Inequality: $ {a\over b} + {b \over c} + {c \over a} \ge 3 $

The $1997$ IIT JEE problem

Proving $(1-a)(1-b)(1-c)(1-d) > 1-a-b-c-d$

Inequality in $L^2$

Show that $x_{n+1} = \frac{x_n + 1}{n+1}$ is decreasing starting from some $n_0$ and find $n_0$.

Proving $\left( 1-\frac{2}{n} \right )^{\frac {n\ln n}{4}}-\left( 1-\frac{1}{n} \right )^{\frac {2n\ln n}{4}}<0$

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