# Questions tagged [rearrangement-inequality]

Proofs of inequalities by using Rearrangement inequality or Chebyshov inequality.

12 questions
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### Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$

I am currently attempting to prove the following inequality $\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$ for all $a,b,c>0$ My instinctive plan of attack is to use the AM/GM ...
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### Does $(\sum_{i=1}^n a_i^{1.5})^2 - \sum_{i=1}^n a_i \; \sum_{i=1}^n a_i a_{i+1} > 0$ hold for $n\le 16$?

Let positive reals $\{a_i\}$, where not all $a_i$ are equal. Does $$f(\{a_i\}) = (\sum_{i=1}^n a_i^{1.5})^2 - \sum_{i=1}^n a_i \; \sum_{i=1}^n a_i a_{i+1} > 0$$ hold for $n \le 16$? It is ...
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### How to prove that $\frac x{\sqrt y}+\frac y{\sqrt x}\ge\sqrt x+\sqrt y$

I am trying to prove that $$\frac x{\sqrt y}+\frac y{\sqrt x}\ge\sqrt x+\sqrt y$$ I tried some manipulations, like multiplications in $\sqrt x$ or $\sqrt y$ or using $x=\sqrt x\sqrt x$, but I'm still ...
### If $a$ and $b$ are positive real numbers such that $a+b=1$, then prove that $\big(a+\frac{1}{a}\big)^2 +\big(b+\frac{1}{b}\big)^2 \ge\frac{25}{2}$ [duplicate]
If $a$ and $b$ are positive real numbers such that $a+b=1$, then prove that $$\bigg(a+ \dfrac {1}{a}\bigg)^2 +\bigg(b+ \dfrac {1}{b}\bigg)^2 \ge \dfrac {25}{2}$$ My tries: I am really unable to see ...
### Prove the inequality $\frac{a^8+b^8+c^8}{a^3b^3c^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ for $a,b,c>0$
As in the title. Prove the inequality $$\frac{a^8+b^8+c^8}{a^3b^3c^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ for $a,b,c>0$. Thsi inequality can be proved in a pretty straightforward manner ...