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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How to prove that $0 < 1$ from the order axioms of $\mathbb R$?

My homework question: From the order axioms for $\mathbb{R}$, show that $0 < 1$. [Hint: From the field axioms, $0 \not=1$. By the trichotomy property, either $0<1$ or $4<0$. Assuming $1 < 0$, get $0 < -1$. Now use Exercise 4.] Exercise 4 from my…
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What is the correct way to compare to algebraic quantities?

If I have two algebraic quantities, what is the correct way to determine if they are equal, or if not which is greater than the other? For example, if I have $\frac{2d}{\sqrt{s^2-c^2}}$ and $\frac{2ds}{s^2-c^2}$, and I want to prove…
JJW5432
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Have I understood square roots in inequalities correctly?

So I've had some trouble understanding what happens when you take the square roots of an inequality. Am I correct in saying that if x^2 > n then -squareroot(n) > x > +squareroot(n) and if x^2 < n then -squareroot(n) > x < +squareroot(n) ??
Jin
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Show that $x+y\geq x^\alpha y^{1-\alpha}$

Is it possible to show that $$x+y\geq x^\alpha y^{1-\alpha}$$ for $\alpha \in(0,1]$ and $x,y\in[0,\infty)$? I tried to manipulate it algebraically, but it does not give me any anything. Equivalently, we need to show that $$x^\alpha - y^\alpha \leq…
Lionville
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Prove or disprove $xyz+\frac{8}{27}\ge xy+yz+zx$ if $x+y+z=1$

if $x,y,z$ are positive and $x+y+z = 1$,Prove:$$xyz+\frac{8}{27}\ge xy+yz+zx$$ Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them. Things I have done: I could rewrite question…
user2838619
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To find maximum value

If $A>0,B>0$ and $C>0$ and further it is known that $A+B+C=\frac{5\pi}{4}$,then find the maximum value of $\sin A+\sin B+\sin C$
Maverick
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How find this $\frac{1}{x-y}+\frac{1}{y-z}+\frac{1}{x-z}$ minimum of the value

let $x,y,z\in R$,and such $x>y>z$,and such $$(x-y)(y-z)(x-z)=16$$ find this follow minimum of the value $$I=\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{x-z}$$ My idea:…
math110
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Prove $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$

if $a,b,c$ are positive real numbers,Prove:$$a^3+b^3+c^3\geq a^2b+b^2c+c^2a$$ Things I have done so far: I know the fact that $$a^3+b^3+c^3\geq\frac{1}{2}[ab(a+b)+bc(b+c)+ca(c+a)]$$ However i tried to reach the problem inequality,but I was not…
user2838619
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How prove this inequality $\sqrt{(2a+1)^2+(2b-\frac{\sqrt{3}}{3})^2}+\sqrt{(2a-1)^2+(2b-\frac{\sqrt{3}}{3})^2}+\cdots$

Question: let $a,b\in R$, show that $$\sqrt{(2a+1)^2+(2b-\dfrac{\sqrt{3}}{3})^2}+\sqrt{(2a-1)^2+(2b-\dfrac{\sqrt{3}}{3})^2}+\sqrt{4a^2+(2b+\dfrac{2\sqrt{3}}{3})^2}\ge…
math110
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How to prove this inequality $b(a-1)(c-1)+c(b-1)(a-1)+a(c-1)(b-1)\le 0$

let $a,b,c>0$,and $$abc=1$$ show that $$b(a-1)(c-1)+c(b-1)(a-1)+a(c-1)(b-1)\le 0$$ since $$b(a-1)(c-1)=b(ac-a-c+1)=abc-ab-bc+b=1-ab-bc+b$$ so we only prove $$3-2(ab+bc+ac)+a+b+c\le 0 $$ oh,this inequality is wrong.let $a=0.1,b=0.1,c=100$,then we…
math110
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Prove the inequality $2\sqrt{x}\ge3-\frac1x $

Given that $x\gt0$ prove the following inequality: $2\sqrt{x}\ge3-\frac1x $ I have done it using calculus but how can I do it using elementary methods?Thanks!!
Kalpan
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if $ax-2by+cz=0$ and $ac-b^2>0$ , Prove $zx-y^2\leq0$

for real numbers like $a,b,c,x,y,z$ that $ax-2by+cz=0$ and $ac-b^2>0$ Prove:$$zx-y^2\leq0$$ Additional info: The Proof should be by contradiction.we can use Cauchy , AM-GM and other simple inequalities. Things I have done so far: as Problem wants…
user2838619
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How prove this ineuality$\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1}\le\sqrt{xy+yz+zx+9}$

let $$x,y,z\in(-1,1), x+y+z=-xyz$$ show that $$\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1}\le\sqrt{xy+yz+zx+9}$$ This problem is my frends ask me,I remenber this is old inequality,But Now I can't it $$x+y+z+3+2\sum_{cyc}\sqrt{xy+x+y+1}\le xy+yz+xz+9$$
math110
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How to prove this inequality $\sum_{i=1}^{n}\left(x^k_{i}\ln{x_{i}}\ln{\frac{x_{i}}{n}}\right)\le 0$

Let $x_{i}\ge 0$ for $i\in\{1,2,\cdots,n\}$ and $x_{1}+x_{2}+\cdots+x_{n}=n$ for $n\ge 3$ Show that for all strictly positive integers $k\ge2$ the following inequality holds : $$\sum_{i=1}^{n}x^k_{i}\ln{x_{i}}\ln{\dfrac{x_{i}}{n}}\le 0$$ We…
math110
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