Question

How do I show this $xy\leq \frac{1}{p}x^p+\frac{1}{q}y^q.$ without using concavity of log function?

score 587 · Answer 1 · 2018-12-26 22:42:18

587

Views

How do I show this $xy\leq \frac{1}{p}x^p+\frac{1}{q}y^q.$ without using concavity of log function?

Published on 26 Dec 2018 - 22:42

score 54 · Answer 2 · 2018-12-27 10:24:21

54

Views

Prove $\int_Rfg\,dm\leq\|f\|_p^{1-p/r}\|g\|_p^{1-q/r}(\int_Rf^pg^q\,dm)^{1/r}$, where $1\leq p\leq\infty$ and $\frac1{r}=\frac1{p}+\frac1{q}-1$

Published on 27 Dec 2018 - 10:24

score 167 · Answer 3 · 2018-12-29 01:19:19

167

Views

Inequality with $(x+y)(y+z)(z+w)(w+x)=1$

Published on 29 Dec 2018 - 1:19

score 61 · Answer 4 · 2019-01-02 08:13:14

61

Views

A Cauchy-Schwarz-type inequality for $\int\prod_n|f_n|$

Published on 02 Jan 2019 - 8:13

score 91 · Answer 5 · 2019-01-06 01:50:18

91

Views

Interchanging limit and integral.

Published on 06 Jan 2019 - 1:50

score 220 · Answer 6 · 2019-01-08 15:07:51

220

Views

Prove that $\frac{a}{c\sqrt{a^2+1}}+\frac{b}{a\sqrt{b^2+1}}+\frac{c}{b\sqrt{c^2+1}}\ge \frac{3}{2}$

Published on 08 Jan 2019 - 15:07

score 190 · Answer 7 · 2026-03-22 14:45:54

190

Views

Alternative form for Liapunov inequality

Published on 22 Mar 2026 - 14:45

score 587 · Answer 8 · 2019-01-11 13:05:53

587

Views

Use of Holder inequality in gradient estimate for harmonic function.

Published on 11 Jan 2019 - 13:05

score 64 · Answer 9 · 2019-01-11 23:43:39

64

Views

Doubt in variant of Holder inequality

Published on 11 Jan 2019 - 23:43

score 133 · Answer 10 · 2019-01-18 05:07:08

133

Views

Cauchy Schwartz Inequality Question: $(a^2+b^2)^3=c^2+d^2 \implies \frac{a^3}{c}+\frac{b^3}{d}\geq 1$

Published on 18 Jan 2019 - 5:07

score 160 · Answer 11 · 2019-01-21 16:11:20

160

Views

Trig Integral Inequality

Published on 21 Jan 2019 - 16:11

score 159 · Answer 12 · 2026-03-24 23:44:57

159

Views

Show that $a^{2014}+b^{2014}\geq a^{2013}+b^{2013} $.

Published on 24 Mar 2026 - 23:44

score 455 · Answer 13 · 2019-02-03 15:30:06

455

Views

Showing that an integral operator on $L^p$ spaces has a certain norm

Published on 03 Feb 2019 - 15:30

score 253 · Answer 14 · 2019-02-07 21:36:19

253

Views

$\frac{1}{3}\sum_{cyc}\frac{1}{\sqrt{1+a}}\ge\frac{1}{\sqrt{1+\sqrt[3]{abc}}}$ if $\;a, b, c\;$ are positive reals s.t. $abc\ge2^9$

Published on 07 Feb 2019 - 21:36

score 258 · Answer 15 · 2019-02-10 12:36:23

258

Views

A lower bound for Holder's inequality and an upper bound for the reverse Holder inequality!

Published on 10 Feb 2019 - 12:36

List Question

How do I show this $xy\leq \frac{1}{p}x^p+\frac{1}{q}y^q.$ without using concavity of log function?

Prove $\int_Rfg\,dm\leq\|f\|_p^{1-p/r}\|g\|_p^{1-q/r}(\int_Rf^pg^q\,dm)^{1/r}$, where $1\leq p\leq\infty$ and $\frac1{r}=\frac1{p}+\frac1{q}-1$

Inequality with $(x+y)(y+z)(z+w)(w+x)=1$

A Cauchy-Schwarz-type inequality for $\int\prod_n|f_n|$

Interchanging limit and integral.

Prove that $\frac{a}{c\sqrt{a^2+1}}+\frac{b}{a\sqrt{b^2+1}}+\frac{c}{b\sqrt{c^2+1}}\ge \frac{3}{2}$

Alternative form for Liapunov inequality

Use of Holder inequality in gradient estimate for harmonic function.

Doubt in variant of Holder inequality

Cauchy Schwartz Inequality Question: $(a^2+b^2)^3=c^2+d^2 \implies \frac{a^3}{c}+\frac{b^3}{d}\geq 1$

Trig Integral Inequality

Show that $a^{2014}+b^{2014}\geq a^{2013}+b^{2013} $.

Showing that an integral operator on $L^p$ spaces has a certain norm

$\frac{1}{3}\sum_{cyc}\frac{1}{\sqrt{1+a}}\ge\frac{1}{\sqrt{1+\sqrt[3]{abc}}}$ if $\;a, b, c\;$ are positive reals s.t. $abc\ge2^9$

A lower bound for Holder's inequality and an upper bound for the reverse Holder inequality!

Trending Questions

Popular # Hahtags

Popular Questions