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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How prove this $ b^2c^2+abc(b+c)+a(b^3+c^3)+a^3(b+c)\ge 2a^2(b^2+c^2+bc)$

let $a,b,c\ge 0$, show that $$ b^2c^2+abc(b+c)+a(b^3+c^3)+a^3(b+c)\ge 2a^2(b^2+c^2+bc)$$ my idea use the SOS methods, But I don't work at last. Thank you
math110
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How prove this inequality $(a^3+b^3+c^3)(ab+bc+ac)\ge 6abc(a^2+b^2+c^2-ab-bc-ac)$

let $a,b,c$ are postive numbers, show that $$(a^3+b^3+c^3)(ab+bc+ac)\ge 6abc(a^2+b^2+c^2-ab-bc-ac)$$ my try: let $$a+b+c=p,ab+bc+ac=q,abc=r$$ and the $$a^3+b^3+c^3=(a+b+c)^3-3(ab+bc+ac)(a+b+c)=p^3-3pq$$ Thank you
math110
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Prove that $4(a+b+c)\ge 3(abc+3); \forall a,b,c > 0: a^2+b^2+c^2=a^2b^2+b^2c^2+c^2a^2.$

Prove that $$4(a+b+c)\ge 3(abc+3); \forall a,b,c> 0: a^2+b^2+c^2=a^2b^2+b^2c^2+c^2a^2.$$ I've tried $pqr$ method. Let $p=a+b+c;q=ab+bc+ca;r=abc.$ The condition gives $$r=\frac{q^2+2q-p^2}{2p}$$ Now, we rewrite the inequality as $$4p\ge 9+3\cdot…
Hello world
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Upper and lower bounds for $f(t) = \int_{-\infty}^\infty \frac{t^2 e^{-t^2/2}}{\cosh(tz)} \phi(z) \, dz$

Consider the function $$f(t) = \int_{-\infty}^\infty \frac{t^2 e^{-t^2/2}}{\cosh(tz)} \phi(z) \, dz.$$ Above $\phi(z)$ is the Gaussian probability density function. I am interested in sharp upper and lower bounds for $f$. I started by using the…
Drew Brady
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$a,b,c>0: \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3.$ Prove that: $\sum_{cyc}\frac{a}{b^2+c^2}\ge\frac{3}{2}\left(\frac{a+b+c}{ab+bc+ca}\right)^2$

Problem. Let $a,b,c>0: \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3.$ Prove that: $$\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\ge\frac{3}{2}\left(\frac{a+b+c}{ab+bc+ca}\right)^2$$ I solved the problem, but my proof is much calculating work.…
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How prove this $\tan{x}>\frac{3x}{2+\sqrt{1-x^2}}$

let $0\dfrac{3x}{2+\sqrt{1-x^2}}$$ This problem have nice solution? my idea: let $$f(x)=\tan{x}-\dfrac{3x}{2+\sqrt{1-x^2}}=\tan{x}-3x\dfrac{2-\sqrt{1-x^2}}{3+x^2}$$ and other idea: let…
math110
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Prove the Inequality: $\sum\frac{x^3}{2x^2+y^2}\ge\frac{x+y+z}{3}$

Let $x, y, z>0$. Prove that: $$\frac{x^3}{2x^2+y^2}+\frac{y^3}{2y^2+z^2}+\frac{z^3}{2z^2+x^2}\ge\frac{x+y+z}{3}$$
my_melody
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Let $a,b,c>0$ and $ab+bc+ca=3$. Prove that $abc(a+b)(b+c)(c+a)\leq8$

Let $a,b,c>0$ and $ab+bc+ca=3$. Prove that $abc(a+b)(b+c)(c+a)\leq8$ My try: $a^3b^3c^3(a+b)(b+c)(c+a)\leq8a^2b^2c^2$ so $(a^2bc+ab^2c)(ab^2c+abc^2)(abc^2+a^2bc)\leq 8ab.bc.ca$. Then choose $ab=x,bc=y,ca=z$. Problem become $x,y,z>0$, $x+y+z=3$.…
tompi2394
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Verifying the triangle inequality

I am going through some practice problems in Abbott's Analysis text, and one of them is the following: Verify the triangle inequality in the special case where a) a and b have the same sign; b) $a \geq 0$, $b < 0$, and $a+b \geq 0$. I do not know…
confused
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Find $\left(\sum_{i=1}^{n}x_{i}\right)\left(\sum_{i=1}^{n}x^3_{i}\right)$ maximum and minimum on the unit $n$-sphere

Let $x_{i}\in \mathbb R$ $(i=1,2,\dots,n)$, $S_{k}=x^k_{1}+x^k_{2}+\cdots+x^k_{n}$, and $S_{2}=1$. Find the maximum and minimum of $S_{1}S_{3}$. If $x_{i}>0$, then using Cauchy-Schwarz inequality we have $$S_{1}S_{3}\ge (S_{2})^2=1.$$ But for real…
math110
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Given $a,b,c>0$, $a+b+c=ab+bc+ca$, prove $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-\sqrt{abc}\ge 2$

Given $a,b,c>0$, $$a+b+c=ab+bc+ca$$, prove $$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-\sqrt{abc}\ge 2$$ I tried with derivatives but haven't solved it yet. Is there a more natural and elementary proof? My progress with derivatives: First $a+b+c=ab+bc+ca\le…
athos
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Inequalities $x/y > 0$

I am sure I am just being stupid but I can't get the illogic in this problem: $$ \begin{align} \frac{x}{y} &> 0\\ \frac{x}{y}\cdot y &> 0\cdot y\\ x &> 0 \end{align} $$ However: when $x= -1,\; y= -1$ ; $-1/-1 > 0$ is true but $x = -1 > 0$…
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Cauchy Schwarz Inequality for numbers

The CS inequality is given by $$x_1y_1 + x_2y_2 \leq \sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}$$ I read that if $x_1 = cy_1$ and $x_2 = cy_2$, then equality holds. But I reduced the above to $c \leq \sqrt{c^2} = |c|$. So isn't this only true if $c…
Lemon
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