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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Prove the inequality.Let a, b and c be nonnegative real numbers.

Let $a$, $b$ and $c$ be nonnegative real numbers. Prove that $a^4+b^4+c^2\ge 8^{½}abc$
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How to prove this general inequality $\displaystyle a\left(\frac{\sin{x}}{x}\right)^m+b\left(\frac{\tan{x}}{x}\right)^n>a+b$

If $$m,n<0,\;a,b>0,\;a\left[\left(\dfrac{2}{\pi}\right)^m-1\right]\ge b,\;am\le 2bn$$ show that $$a\left(\dfrac{\sin{x}}{x}\right)^m+b\left(\dfrac{\tan{x}}{x}\right)^n>a+b,\qquad\forall x\in\left(0,\dfrac{\pi}{2}\right)$$ I know this is Wilker…
math110
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Inequality with two moduli

I have a question similar to this, find all $x \in \mathbb{R}$ satisfying $\displaystyle 3 < \left| x+1 \right| + \left| x - \frac{1}{2} \right| < 7$ which is rather trivial by distinguishing cases, the question suggests an alternative method which…
suarko
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If positive $a,b,c,d$ satisfy $(a^3+b^3)^4=c^3+d^3$, prove that $a^4c+b^4d\ge cd$.

If positive $a,b,c,d$ satisfy $(a^3+b^3)^4=c^3+d^3$, prove that $$a^4c+b^4d\ge cd$$ It kind of seems useful to begin with a division of both sides by $cd$: $$\frac{a^4}{d}+\frac{b^4}{c}\ge1$$ It seems like a simple Cauchy-Schwarz would be of use…
user26486
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Help on this inequality

If a,b,c are positive numbers, prove the inequality $$ \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} ≥ \frac{3}{1+abc} $$
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Prove that $a^n - b^n + c^n - d^n \ge (a-b+c-d)^n$

Following on from an earlier question, and in search of a conceptual insight, I asked myself: Given real numbers $a \ge b \ge c \ge d \ge 0 \tag{1}$ Prove that $a^n - b^n + c^n - d^n \ge (a - b + c - d)^n \text{ for all } \underline{n \in…
Anant
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If $\sum x_k= \sqrt{n}$, then $\sum\frac{x_k^2}{(x_k^2+1)^2}\leq\frac{n^2}{(n+1)^2}$

A new question has emerged after this one was successfully answered by r9m: If $a+b+c+d = 2$, then $\frac{a^2}{(a^2+1)^2}+\frac{b^2}{(b^2+1)^2}+\frac{c^2}{(c^2+1)^2}+\frac{d^2}{(d^2+1)^2}\le \frac{16}{25}$. I thought of this generalization. Does it…
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How prove this inequality $\sum\limits_{cyc}\frac{1-2\sin{\frac{C}{2}}}{\sin{\frac{B}{2}}}\ge 0$

in $\Delta ABC$,prove or disprove $$\dfrac{1-2\sin{\dfrac{C}{2}}}{\sin{\dfrac{B}{2}}}+\dfrac{1-2\sin{\dfrac{A}{2}}}{\sin{\dfrac{C}{2}}}+\dfrac{1-2\sin{\dfrac{A}{2}}}{\sin{\dfrac{A}{2}}}\ge 0$$ My idea:…
math110
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Inequality: $x^2+y^2+z^2 \geq \sqrt{2}x(z+y)$

How can I prove the following inequality: $$x^2+y^2+z^2 \geq \sqrt{2}x(z+y)?$$ Thanks!
Iuli
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Subtracting positive numbers from denominator in an inequality.

When we have the following inequality: $$\frac{a}{b+c} \ge \frac{d}{e+c},$$ with $a,b,c,d,e \in \mathbb R_{\ge 0}$ Then it seems to hold that $$\frac{a}{b} \ge \frac{d}{e},$$ Is this correct? Does it also work in the other direction (iff)?
miselico
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Find Max value of this expression: $P=(x-2yz)(y-2zx)(z-2xy)$

Let $x$, $y$ and $z$ be positives and satisfying $x^2+y^2+z^2=2xyz+1$ . Find a maximum of this expression: $$P=(x-2yz)(y-2zx)(z-2xy).$$
abcdxyz
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Inequality $\sum_{cyc}\frac{(a+b)^{3}}{\sqrt[3]{2(a+b)(a^{2}+b^{2})}}\ge{12}$ when $ab+bc+ca=6$

Let $a$, $b$ and $c$ be non-negative reals such that $ab+bc+ca=6$. Prove that: $$\frac{(a+b)^{3}}{\sqrt[3]{2(a+b)(a^{2}+b^{2})}}+\frac{(b+c)^{3}}{\sqrt[3]{2(b+c)(b^{2}+c^{2})}}+\frac{(c+a)^{3}}{\sqrt[3]{2(c+a)(c^{2}+a^{2})}}\ge{12}$$
user136728
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Prove inequality and when does equality hold?

Prove that for any real positive numbers $a$ and $b$ $$\left(a+\frac{1}{b}\right)\left(b+\frac{1}{a}\right) \geq 4.$$ When does equality hold?
Tim
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Solving an equality in 2 variables

I need to prove that $$\left(a + \frac{1}{a}\right)^2 +\left(b + \frac{1}{b}\right)^2 \gt \frac{25}{2}$$ if $a+b = 1$ and $a b \le 1/4$ I'd like a hint. Solve the equality first to $a$ or $b$, or stay in a and b as to get $a b \le 4$ in the…
Ignace
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$ \dfrac a{\sqrt[4] {16+b-d}} +\dfrac b{\sqrt[4] {16+c-a}}+ \dfrac c{\sqrt[4] {16+d-b}}+\dfrac d{\sqrt[4] {16+a-c}} \ge 8$

Let $a,b,c,d>0 ; a+b+c+d=16$ , then how to prove that $ \dfrac a{\sqrt[4] {16+b-d}} +\dfrac b{\sqrt[4] {16+c-a}}+ \dfrac c{\sqrt[4] {16+d-b}}+\dfrac d{\sqrt[4] {16+a-c}} \ge 8$ ?
Souvik Dey
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