Questions tagged [inequality]

Questions on proving, manipulating and applying inequalities. Do not use this tag just because an inequality appears somewhere in your question.

An inequality is a mathematical relation between two quantities that are not necessarily equal, but bigger or smaller.

To prove inequalities, a number of proven inequalities can be used, including:

  • The AM-GM inequality

    Let $x_i>0$, $\alpha_i>0$ such that $\alpha_1+\alpha_2+...+\alpha_n=1$. Prove that $$\alpha_1x_1+\alpha_2x_2+...+\alpha_nx_n\geq x_1^{\alpha_1}x_2^{\alpha_2}...x_n^{\alpha_n}$$

For $\alpha_1=\alpha_2=...=\alpha_n=\frac{1}{n}$ we obtain the well-known $$\frac{x_1+x_2+\cdots+x_n}{n} \ge \sqrt[n]{x_1x_2\cdots x_n}$$

  • The Power Mean inequality (P-M).

    Let $a_1, a_2,\cdots, a_n$ be positive numbers and $p>q$. Then $$\left(\frac{a_1^p+a_2^p+\cdots+a_n^p}{n}\right)^{\frac{1}{p}} \geq \left(\frac{a_1^q+a_2^q+\cdots+a_n^q}{n}\right)^{\frac{1}{q}}$$

  • The Rearrangement inequality (R).

    Let $a_1\le\dots\le a_n$ and $b_1\le\dots\le b_n$. For all permutations $\sigma\in S_n$, $$\sum_{i=1}^na_ib_{n-i+1}\le\sum_{i=1}^na_ib_{\sigma(i)}\leq\sum_{i=1}^na_ib_i.$$

The rearrangement generalizes similar for more than two sequences of numbers.

  • The Cauchy-Schwarz inequality (C-S).

    If $a_1, a_2, \cdots, a_n$ and $b_1, b_2,\cdots, b_n$ are two sequences of real numbers, then $$\sum^{n}_{i=1} a_i^2 \sum^{n}_{i=1} b_i^2\geq\left(\sum^{n}_{i=1} a_ib_i \right)^2$$

  • The H$\ddot o$lder inequality (H).

    Let $a_1$, $a_2$,..., $a_n$, $b_1$, $b_2$,..., $b_n$, $\alpha$ and $\beta$ be positive numbers. Then $$\left(\sum_{i =1}^n a_i\right )^\alpha \left(\sum_{i =1}^n b_i \right )^\beta\geq \left(\sum_{i =1}^n (a_ib_i)^\frac{1}{\alpha+\beta}\right )^{\alpha+\beta} $$

  • The Schur inequalities (S):

    Let $x$, $y$ and $z$ be positive numbers and $t$ is a real number. Prove that:$$x^t(x-y)(x-z)+y^t(y-z)(y-x)+z^t (z-x)(z-y)\ge 0$$

  • Muirhead inequalities

    A sequence $a_1 \geq a_2 \geq \dots \geq a_n$ majorizes a sequence $b_1 \geq b_2 \geq \dots \geq b_n$ if $$\sum_{i=1}^k a_i \geq\sum_{i=1}^k a_i $$ for all $1\leq k < n$ and $$\sum_{i=1}^n a_i =\sum_{i=1}^n a_i $$ If sequence $(a_i)$ majorizes $(b_i)$ (notated as $a_i \succ b_i$), then $$\sum_{\text{sym}}x_1^{a_1}x_2^{a_2}\dots x_n^{a_n}\geq \sum_{\text{sym}}x_1^{b_1}x_2^{b_2}\dots x_n^{b_n}$$

30160 questions
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Inequalities and multiplication

Given two inequalities the rules of multiplication are said to be (check link for the source): 1.) If $0
Sooraj S
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Can we say that $\left(a+b\right)^{\alpha}>a^{\alpha}+b^{\alpha}$ for all $a,b>0$ and $\alpha>1$?

For $\alpha\in\mathbb{N}$ we can use the Binomial and get: $$\left(a+b\right)^{n}=\sum_{k=0}^{n}{n \choose k}a^{k}b^{n-k}=\sum_{k=1}^{n-1}{n \choose k}a^{k}b^{n-k}+a^{n}+b^{n}>a^{n}+b^{n} $$ But what about rational and irrational…
Jon
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How prove this inequality $(a+b)^rb^{2n-r}\le\dfrac{n}{4}+r,$

let $a,b>0$,and $n\ge 4$ be postive integers, such $(a+b)^{2n}=2n+\dfrac{n}{4},a^{2n}=\dfrac{n}{4}$ show that $$(a+b)^rb^{2n-r}\le\dfrac{n}{4}+r,\forall r\in[0,2n]$$ it seem hard for $n=4$ case.show this inequality $3^{\frac{n}{4}}\cdot…
math110
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How can this cubic inequality be solved? (2015 A-level paper)

I'm trying to complete the following question (from the AQA June 2015 FP1 A-level paper): By first finding a suitable cubic inequality for $k$, find the greatest value of $k$ for which $$\sum_{r=k+1}^{60} (3r+2) \log_8 4^r$$ is greater than…
ash
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How to prove this inequality involving ratio of sums?

Let $K=\{1,\ldots,k\}$ be a set of $k$ jobs. Each job $j\in K$ has a positive weight $w_j$ and a positive profit $p_j$. Let $\ell^*$ be the job such that: $$\dfrac{\sum_{j=1}^{\ell^*}w_j}{\sum_{j=1}^{\ell^*}p_j}=\max\limits_{1\leq\ell\leq…
zdm87
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How would you prove that $2^{50} < 3^{33}$ without directly calculating the values

Could you generalise the question and get something along the lines of $n^{50} < (n+1)^{33}$ ?
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Inequality Proof

Given that $$x>1$$Prove$$\dfrac{x^{\dfrac{1}{\ln 2}}\ln \left(1+\dfrac{1}{x}\right)}{\ln (1+x)}>1$$ I found that $\ln 2$ is the largest number for this inequality to hold via some numerical verifications. However couln't find a way to prove it.…
Mike
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How to prove this inequality? Thanks

Let $x_{1},x_{2},\cdots,x_{n}\in [0,1]$, show that $$\sum_{i=1}^{n}\left[\left(\prod_{j\neq i }x_{j}\right)^{\frac{n}{2}}\cdot\sqrt{1-x_{i}}\right]\le 1$$ Maybe it can use McLaughlin inequality to solve it?I found this inequality simaler this…
math110
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How to prove that $(\cos\theta)^p\leqslant\cos(p\theta)$ for every $0≤p≤1$ and $0≤θ≤π/2$

How can I prove the following inequality: $$\forall p\in[0,1], \theta\in [0,π/2];\cos^p(θ)≤\cos(pθ)$$ Please help me to complete this proof . I suppose $g(θ)=\cos^p(θ)-\cos(pθ) $ and I want show that $g(θ)≤0$. And I use derivative and integral…
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Prove that $e^x>x(x+1)$

Let $x>0$,prove that $$e^x>x(x+1)$$or $$x>\ln{x}+\ln{(x+1)}$$ we can use this Taylor some first four term, $$e^x>1+x+\dfrac{1}{2}x^2+\dfrac{1}{6}x^3$$ prove it But this inequality it seem very nice,maybe there exsit simple methods or Amazing…
math110
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Inverting the RMS-AM inequality

The RMS-AM inequality states that for positive real numbers $x_1,\ldots,x_n$, $$AM=\frac{x_1+\cdots+x_n}{n}\leq\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}}=RMS.$$ For two positive numbers $x_1,x_2$, the inequality can be inferred geometrically from the…
Auslander
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Inequality of arithmetic and geometric mean

I did a proof for inequality below, anyone has a other proof? Let $a$ and $b$ be positive real numbers, and $t$ the parameter. Prove that: $$a+b\geq 2\sqrt{1-t^2}\sqrt{ab}+(a-b)t$$
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Prove that $k(x^6+y^6+z^6)+xyz(x^3+y^3+z^3)\geq0$

Let $x$, $y$ and $z$ be real numbers and $k=\frac{(2+\sqrt7)\sqrt[3]{3-\sqrt7}}{6}$. Prove that: $$k(x^6+y^6+z^6)+xyz(x^3+y^3+z^3)\geq0$$ The equality occurs here for $\frac{x}{\sqrt[3]{\sqrt7-3}}=y=z$. I tried the following way. Let…
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Proof of an inequality using Cauchy's inequality: $\sum_{k=1}^n \left(p_k+\frac{1}{p_k}\right)^2\geq n^3+2n+\frac{1}{n}$

Given that $p_k> 0$ and $p_1+p_2+\cdots+p_n=1$, prove that \begin{equation} \sum_{k=1}^n \left(p_k+\frac{1}{p_k}\right)^2\geq n^3+2n+\frac{1}{n}. \end{equation} I believe that Cauchy's inequality should be used at some point but I haven't figured…