# Questions tagged [rearrangement-inequality]

Inequalities related to the rearrangement inequality or Chebyshev inequality

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### Inequality $a^4 + b^4 + c^4 + k(a^3b + b^3c + c^3a)$ $\ge (k + 1)(ab^3 + bc^3 + ca^3)$ where $k\in [0,1]$

Prove the following inequality for all real numbers $a, b, c$ and for $k \in [0, 1]$: $$a^4 + b^4 + c^4 + k(a^3b + b^3c + c^3a) \geq (k + 1)(ab^3 + bc^3 + ca^3)$$ I have tried to prove by ...
219 views

### distance between sorted arrays

Assume we have two arrays of real numbers: $$X = \{x_{1}, x_{2}, \dots, x_{n} \}$$ and $$Y = \{y_{1}, y_{2}, \dots, y_{n} \}$$ Next, assume that $d = \max(|x_{i} - y_{i}|)$. Next let us sort both ...
• 99
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### Prove: $\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\ge\sqrt{3}\sum_{cyc}{\sqrt[4]{\frac{5ab}{c}+4a}}$

Prove that the following inequality :$$\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\ge\sqrt{3}\left(\sqrt[4]{\frac{5ab}{c}+4a}+\sqrt[4]{\frac{5bc}{a}+4b}+\sqrt[4]{\frac{5ca}{b}+4c}\right)$$ holds for all ...
227 views

### Let $x\geq y\geq z>0$. Prove that $\frac {x^{2}y}{z} + \frac {y^{2}z}{x} + \frac {z^{2}x}{y}\geq x^{2} + y^{2} + z^{2}.$

Let $x\geq y\geq z>0$. Prove that $$\frac {x^{2}y}{z} + \frac {y^{2}z}{x} + \frac {z^{2}x}{y}\geq x^{2} + y^{2} + z^{2}.$$ The problem is from Vietnamese MO 1991 and has been posted here before. ...
• 3,956
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### Let $h_{1},h_{2},h_{3}$ be the altitudes and $m_{1},m_{2},m_{3}$ be the medians of the triangle ABC.

Show that:$$\frac{h_1}{m_1}+\frac{h_2}{m_2}+\frac{h_3}{m_3}\leq3$$ So, I was wondering if we could prevent all the hefty geometry and solve this using Chebyshev's or the Rearrangement inequality. ...
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### Can I solve this blood question by re-arranging? [closed]

I believe the answer to the question in the attached image is the 4th bubble. Be careful not to accidentally miss the formula under the worded explanation, at the bottom of the image. The unit litre/...
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1 vote
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### Given 4 numbers $a, b, c, d> 0,$ show $16\max\limits_{\bigcirc}\left \{ a^{3}+ 3bcd \right \}\!\geq\!\left ( a+ b+ c+ d \right )^{3}$

Given four positive numbers $a, b, c, d.$ Prove that $$16\max\left \{ a^{3}+ 3bcd, b^{3}+ 3cda, c^{3}+ 3dab, d^{3}+ 3abc \right \}\geq\left ( a+ b+ c+ d \right )^{3}$$ the way I think is using the ...
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### With $n> 3,$ prove that $a_{1}+ a_{2}+ a_{3}\geq 100$ by using Karamata's inequality

Given $n$ real numbers $a_{1}, a_{2}\cdots a_{n}$ so that $$a_{1}\geq a_{2}\geq\cdots\geq a_{n}, a_{1}+ a_{2}+ \cdots+ a_{n}= 300, a_{1}^{2}+ a_{2}^{2}+ \cdots+ a_{n}^{2}> 10000$$ With $n> 3,$ ...
1 vote
287 views

### Inequality $a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0$

Let $a,b,c$ be the lengths of the sides of a triangle. Prove that: $$a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0.$$ Now, I am supposed to solve this inequality by applying only the Rearrangement ...
• 2,143
1 vote
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### If $x,y,z>0$, $n,m\in N$, $n\ge m$ prove:$\frac{x^n}{(y+z)^m}+\frac{y^n}{(z+x)^m}+\frac{z^n}{(x+y)^m}\ge \frac{1}{2^m}(x^{n-m}+y^{n-m}+z^{n-m})$

Given positive numbers $x,y,z$ and $n,m$ positive integers with $n\ge m$ prove that: $$\frac{x^n}{(y+z)^m}+\frac{y^n}{(z+x)^m}+\frac{z^n}{(x+y)^m}\ge \frac{1}{2^m}(x^{n-m}+y^{n-m}+z^{n-m})$$ I tried ...
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• 11.9k
1 vote
118 views

### $\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$ Proof

Does anyone know hot to prove this inequality? Having: $a, b, c \gt 0$ $$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$$ I tried with the AM-GM inequality but I couldn't get ...
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### Verification to solution an inequality and proving another.

I need solution verification an inequality, that I have solved because it seems too good to be true. But first, I attempted this but couldn't complete: Let $a$, $b$ and $c$ be the sides of a triangle ...
• 1,516
161 views

### Proving a cyclic inequality

Show that $a^4 + b^4 + c^4 \geq a^3b + b^3c + c^3a$ for any postive integers $a, b, c$ I'm not sure how to approach this problem. I've tried assuming that WLOG $a > b > c$ so that it is clear ...
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• 11.9k
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### prove $\sum_{i=1}^{n}\sqrt{a_i}\ge (n-1)\sum_{i=0}^{n}\frac{1}{\sqrt{a_i}}$ [duplicate]

prove $$\sum_{i=1}^{n}\sqrt{a_i}\ge ({n-1})\sum_{i=1}^{n}\frac{1}{\sqrt{a_i}}$$ if $$\sum_{i=1}^{n}\frac{1}{1+a_i}=1$$ My try: i tried substituiting $y_i=\frac{1}{1+a_i}$ thus $\sum y_i=1$ also ...
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### prove $\sum_{cyc}\frac{a^3}{b}\ge ab+bc+ca$ if $a,b,c>0$

prove $\sum_\text{cyc}\frac{a^3}{b}\ge ab+bc+ca$ if $a,b,c>0$ I couldn't proceed much. I tried rearranging the inequality and it became $a^4c+b^4a+c^4b\ge a^2b^2c+b^2c^2a+c^2a^2b.$ I tried using ...
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### Rearrangement Inequality- $\sum_{c y c} \frac{a^{2}+b c}{b+c} \geq a+b+c$

$$\sum_{c y c} \frac{a^{2}+b c}{b+c} \geq a+b+c$$ I'm confused about how to solve this. can someone give me a few hints? I'm stuck thing what even $\sum_{c y c}$ means!
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### Is this Factorization?

I'm doubtful about the some parts of the solution to this question: Suppose that the real numbers $a, b, c > 1$ satisfy the condition $${1\over a^2-1}+{1\over b^2-1}+{1\over c^2-1}=1$$ Prove ...
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### If $a^2 + b^2 + c^2 = 1$, what is the the minimum value of $\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}$?
Suppose that $a^2 + b^2 + c^2 = 1$ for real positive numbers $a$, $b$, $c$. Find the minimum possible value of $\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}$. So far I've got a minimum of $\sqrt {3}$...