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 $$\sum_{\text{sym}}x_1^{a_1}x_2^{a_2}\dots x_n^{a_n}\geq \sum_{\text{sym}}x_1^{b_1}x_2^{b_2}\dots x_n^{b_n}$$

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Ineqality problem

My mentor gave this problem to our class. later on he also told the solution to the problem.but i did not understood the solution as it was done in a very complicated way. so can anyone help me the solve this problem with proper…
Marble
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Inequality $\frac{x^3+1}{\sqrt{x^4+y+z}}+\frac{y^3+1}{\sqrt{y^4+z+x}}+\frac{z^3+1}{\sqrt{z^4+x+y}}\geq 2\sqrt{3}$

If we Let $x,y,z>0$ such that $x+y+z=3$. how to prove that $$\frac{x^3+1}{\sqrt{x^4+y+z}}+\frac{y^3+1}{\sqrt{y^4+z+x}}+\frac{z^3+1}{\sqrt{z^4+x+y}}\geq 2\sqrt{3}$$
abf
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Prove that $\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{c+a}{ca}}$

For $a,b,c$ are positive real number. Prove that $$\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{c+a}{ca}}$$ Let…
Word Shallow
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Starting an inequality family: How far can we go?

Consider the inequality $$(ab+bc+ca)\cdot\left(\frac{1}{(ka+b)^2}+\frac{1}{(kb+c)^2}+\frac{1}{(kc+a)^2}\right)\,\ge\,\frac{9}{(k+1)^2}\tag{"Case $k$"}$$ with variables $\,a,b,c\in\mathbb{R}^{>0}\,$ and parameter $\,k\in\mathbb{R}$. (At least) the…
Hanno
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Real numbers $a$, $b$, $c$ and $d$ satisfy the inequality $abcd > a^2 + b^2 + c^2 + d^2$. Prove that $abcd > a + b + c + d + 8$.

Real numbers $a$, $b$, $c$ and $d$ satisfy the inequality $abcd > a^2 + b^2 + c^2 + d^2$. How to prove that $abcd > a + b + c + d + 8$? I've tried using AM-GM inequality but with no outcome. The problem was taken from 57-th Belarusian Mathematical…
user4201961
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Inequality with 4 variables, powers.

$\{ a,b,c,d \in\Bbb R_+\ \}$ $a^6 + b^3 + c^2 + d \ge 2 \sqrt[3]{2} \sqrt[4]{3} \sqrt {abcd} $ I have no idea how to do this sort of inequalities, never seen it before.
VereX
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Prove that $x<2^x$

How can we prove that given that $x\leq 2$, then $2^x>x$? I think that this seems to be intuitively correct but I don't know how to prove it. Can it be proven without calculus?
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determinant inequality

Given integers a,b,c and d; is the absolute value of the determinant of [a c;b d] less than or equal to the absolute value of [a e;b f] + absolute value of [e c;f d] for all a,b,c,d? In other words, is |ad-bc| less than or equal to |af-be| + |de-cf|…
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Positive $x,y,z$, prove $\frac{(x^2+y^2+z^2)^2}{x^3y+y^3z+z^3x} \geq 2 (\frac{xy^2+yz^2+zx^2}{x^2y+y^2z+z^2x})+\frac{x^2y+y^2z+z^2x}{xy^2+yz^2+zx^2}$

$x,y,z \geqslant 0$, prove $$\frac{(x^2+y^2+z^2)^2}{x^3y+y^3z+z^3x} \geqslant 2 \left(\frac{xy^2+yz^2+zx^2}{x^2y+y^2z+z^2x} \right)+\frac{x^2y+y^2z+z^2x}{xy^2+yz^2+zx^2}$$ What I tried Denote $u=xy^2+yz^2+zx^2$ and $v=x^2y+y^2z+z^2x$. There is a…
HN_NH
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Given the $ x+y+z =3$ Prove that $ x^3+y^3+z^3+6xyz \geq 27/4$

Also, $x$,$y$ and $z$ are positive. I have figured out that it would be enough to prove that LHS is bigger than $(27-\text{LHS})/3$, then by substituting 27 with $ (x+y+z)^3 $, it's enough to prove that $$x^3+y^3+z^3+6xyz \geq…
TAJD
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Prove the following inequality $x^2+y^2+1>x\sqrt{y^2+1}+y\sqrt{x^2+1}$

Can anybody help me prove this inequality ?
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An inequality with a load of variables: $ (1-a_1)(1-a_2)...(1-a_n) \ge 1/2$

It is known about numbers $$a_1, a_2, ... a_n$$ that $$a_1 + a_2 +...+a_n \le 1/2.$$ Prove that $$ (1-a_1)(1-a_2)...(1-a_n) \ge 1/2$$ I have tried using $a^2 \geq 0$, it led to nothing. How can I make my inequality look like in the possible…
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Show $\forall a,b\in\mathbb{R}^*\quad 3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2 (\frac{a}{b}+\frac{b}{a})$

Let $a,b \in \mathbb{R}^*$ Show that : $$3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2\left(\frac{a}{b}+\frac{b}{a}\right)$$ Let $a,b\in\mathbb{R}^*$ let $t=\dfrac{a}{b}+\dfrac{b}{a}$ , we've : $$ \begin{aligned} 3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge…
Educ
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Prove that $\left(x_1+\dots+x_k\right)^j\leq k^{j-1}\left(x_1^j+\dots+x_k^j\right)$ for $x_1,\dots,x_k\geq0$.

Let $j,k\in\mathbb N$, with $j,k\geq 1$. Prove that $$\left(x_1+\dots+x_k\right)^j\leq k^{j-1}\left(x_1^j+\dots+x_k^j\right)$$ for $x_1,\dots,x_k\geq0$. A proof by induction seems very difficult.
Mark
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