Questions tagged [holder-inequality]

Proving or manipulations with inequalities by using Hölder's inequality.

  • Let $a_1$, $a_2$,..., $a_n$, $b_1$, $b_2$,..., $b_n$, $\alpha$ and $\beta$ be positive numbers. Prove that: $$(a_1+a_2+...+a_n)^{\alpha}(b_1+b_2+...+b_n)^{\beta}\geq$$$$\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}+...+\left(a_n^{\alpha}b_n^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}$$

  • Let $a_{ij}>0$ and $\alpha_i>0$. Prove that: $$\prod_{i=1}^k\left(\sum_{j=1}^na_{ij}\right)^{\alpha_i}\geq\left(\sum_{j=1}^n\left(\prod_{i=1}^ka_{ij}^{\alpha_i}\right)^{\frac{1}{\sum\limits_{i=1}^k\alpha_i}}\right)^{\sum\limits_{i=1}^k\alpha_i}$$

  • Let $f$ and $g$ be positive integrable functions on $[a,b]$ and let $\alpha$ and $\beta$ be positive numbers. Prove that: $$\left(\int\limits_{a}^bf(x)dx\right)^{\alpha}\left(\int\limits_{a}^bg(x)dx\right)^{\beta}\geq\left(\int\limits_{a}^b\left(f(x)^{\alpha}g(x)^{\beta}\right)^{\frac{1}{\alpha+\beta}}dx\right)^{\alpha+\beta}$$

591 questions
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On the equality case of the Hölder and Minkowski inequalities

I'm following the book Measure and Integral of Richard L. Wheeden and Antoni Zygmund. This is problem 4 of chapter 8. Consider $E\subseteq \mathbb{R}^n$ a measurable set. In the following all the integrals are taken over $E$, $1/p + 1/q=1$, with…
16
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Reverse Holder Inequality $\|fg\|_1\geq\| f\|_{\frac{1}{p}}\|g\|_{-\frac{1}{p-1}}$

Let $p\in(1,\infty)$ and $(X,\mathcal{F},\mu)$ a measure space such that $\mu(X)\not=0$. Let $f,g:X\to\mathbb{R}$ be such that $g\not=0$ a.e., $\|fg\|_1<\infty$ and $\|g\|_{-\frac{1}{p-1}}<\infty$. Then prove:$$\|fg\|_1\geq\|…
Marcelo
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16
votes
6 answers

Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$

If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn't help(I got the reverse inequality). Need some stronger…
14
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Prove $(a_1+b_1)^{1/n}\cdots(a_n+b_n)^{1/n}\ge \left(a_1\cdots a_n\right)^{1/n}+\left(b_1\cdots b_n\right)^{1/n}$

consider positive numbers $a_1,a_2,a_3,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$. does the following in-equality holds and if it does then how to prove it $\left[(a_1+b_1)(a_2+b_2)\cdots(a_n+b_n)\right]^{1/n}\ge \left(a_1a_2\cdots…
Mia
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13
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3 answers

Proving $\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$

The inequality: $$\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$$ Conditions: $a,b,c,d \in \mathbb{R^+}$ I tried using the normal Cauchy-Scharwz, AM-RMS,…
13
votes
6 answers

Prove $\sum\limits_{\mathrm{cyc}} \sqrt{a^3+2} \ge \sqrt{\frac{9+3\sqrt{3}}{2}(a^2+b^2+c^2)}$ for $a, b, c \ge 0$

Let $a,b,c\ge 0$,show that $$\sqrt{a^3+2}+\sqrt{b^3+2}+\sqrt{c^3+2}\ge \sqrt{\dfrac{9+3\sqrt{3}}{2}(a^2+b^2+c^2)}$$
math110
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How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$

How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$ for $a,b,c>0$ and $abc=1$? I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}{4\sqrt{2}}$ Indeed,$\frac{{{a}^{2}}}{1+a}\ge…
12
votes
1 answer

Minimal order of $\bigg|\sum\limits_{k=1}^na^3_k\bigg|$ given $\bigg|\sum\limits_{k=1}^na_k\bigg|\geq\sqrt{n}$

Does there exist a sequence of real numbers $\{a_k\}_{k=1}^\infty$ such that for every $n\in\mathbb{N}$ we have $$\bigg|\sum\limits_{k=1}^na_k\bigg|\geq\sqrt{n}$$ but $$\bigg|\sum\limits_{k=1}^na^3_k\bigg|\leq \frac{c}{n}$$ for some constant $c>0$…
Aron
  • 197
12
votes
2 answers

If $u \notin L^p[0,1]$, can we find some $w \in L^{\frac{p}{p-1}}[0,1]$ such that $\int_0^1 u \cdot w = \infty$?

The question is as in the title. For some fixed $p \in (1,\infty)$, let $p' \in (1,\infty)$ be such that $\frac{1}{p}+\frac{1}{p'}=1$. Then, for any $u \in L^1[0,1] - L^p[0,1]$, I wonder if it is possible to find some $w \in L^{p'}[0,1]$ such…
12
votes
3 answers

Prove that $({a\over a+b})^3+({b\over b+c})^3+ ({c\over c+a})^3\geq {3\over 8}$

Let $a,b,c$ be positive real numbers. Prove that $$\Big({a\over a+b}\Big)^3+\Big({b\over b+c}\Big)^3+ \Big({c\over c+a}\Big)^3\geq {3\over 8}$$ If we put $x=b/a$, $y= c/b$ and $z=a/c$ we get $xyz=1$ and $$\Big({1\over 1+x}\Big)^3+\Big({1\over…
12
votes
2 answers

$\sum\limits_{i=1}^n \frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod \limits_{j=1}^nx_j}} \ge 1$, for all $x_i>0.$

Can you prove the following new inequality? I found it experimentally. Prove that, for all $x_1,x_2,\ldots,x_n>0$, it holds that $$\sum_{i=1}^n\frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod\limits _{j=1}^nx_j}} \ge 1\,.$$
12
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3 answers

Minimum of $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}$

What is the minimum of $$f(a,b,c):=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}$$ where $a,b,c$ are positive real numbers? When $a=b=c$, we have $f(a,b,c)=\dfrac{3}{\sqrt{2}}\approx 2.12$ When $a=1,b=c\rightarrow\infty$, we have…
11
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5 answers

Prove that $\frac{m}{\sqrt{7m+2}} + \frac{n}{\sqrt{7n+2}} + \frac{p}{\sqrt{7p+2}} \ge 1.$

Let $m,n,p \ge 0$ such that $m+n+p = mn+np+pm > 0$. Prove that: $$\frac{m}{\sqrt{7m+2}} + \frac{n}{\sqrt{7n+2}} + \frac{p}{\sqrt{7p+2}} \ge 1.$$ The equality occurs when $(m,n,p)=(1,1,1)$ or $(m,n,p)=(2,2,0)$ I tried to use Holder…
10
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3 answers

An Inequality Problem $1 \le \frac{a}{1-ab}+\frac{b}{1-bc}+\frac{c}{1-ac} \le \frac{3\sqrt{3}}{2}$

If $a,b,c>0$, are positive real numbers such that $a^2+b^2+c^2=1$ then, the following Inequalities hold: $\displaystyle 1 \le \frac{a}{1-ab}+\frac{b}{1-bc}+\frac{c}{1-ac} \le \frac{3\sqrt{3}}{2}$ $\displaystyle 1 \le…
r9m
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Proving $\sum\limits_{\text{cyc}} \frac{a}{b^2+c^2+d^2} \geq \frac{3\sqrt{3}}{2}\frac{1}{\sqrt{a^2+b^2+c^2+d^2}}$ for $a, b, c, d >0$

Let $a, b, c, d > 0$. Prove that $$\frac{a}{b^2+c^2+d^2}+\frac{b}{a^2+c^2+d^2}+\frac{c}{a^2+b^2+d^2}+\frac{d}{a^2+b^2+c^2}$$ $$\geq\frac{3\sqrt{3}}{2}\frac{1}{\sqrt{a^2+b^2+c^2+d^2}}.$$ What I tried, was to say that $a^2+b^2+c^2+d^2=1$ and…
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