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Questions tagged [rearrangement-inequality]

Proofs of inequalities by using Rearrangement inequality or Chebyshov inequality.

9
votes
5answers
542 views

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 ...
14
votes
2answers
369 views

Prove that $\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}+\cdots+\frac{1}{\sqrt{x_n}} \right )$

Let $x_1,x_2,\ldots,x_n > 0$ such that $\dfrac{1}{1+x_1}+\cdots+\dfrac{1}{1+x_n}=1$. Prove the following inequality. $$\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\dfrac{1}{\sqrt{...
4
votes
3answers
171 views

Find minimum value of $\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$

If $a,b,c$ are sides of triangle Find Minimum value of $$S=\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$$ My Try: Let $$P=\sqrt{a}+\sqrt{b}+\sqrt{c}$$ we have $$S=\sum \frac{1}{\frac{\sqrt{b}}{...
2
votes
0answers
101 views

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 ...
10
votes
2answers
415 views

Proving the inequality $4\ge a^2b+b^2c+c^2a+abc$

So, a,b,c are non-negative real numbers for which holds that $a+b+c=3$. Prove the following inequality: $$4\ge a^2b+b^2c+c^2a+abc$$ For now I have only tried to write the inequality as $$4\left(\frac{...
4
votes
4answers
138 views

Algebraic proof that if $a>0$ then $1+a^9 \leq \frac{1}{a}+a^{10} $

Prove that if $a>0$ then $1+a^9 \leq \frac{1}{a}+a^{10} $. Using the fact that $a \gt 0$, multiply by $a$ on both sides and get everything to one side we have; $a^{11}-a^{10}-a+1 \geq 0$. By ...
3
votes
4answers
585 views

Inequality: $ab^2+bc^2+ca^2 \le 4$, when $a+b+c=3$.

Let $a,b,c $ are non-negative real numbers, and $a+b+c=3$. How to prove inequality $$ ab^2+bc^2+ca^2\le 4.\tag{*} $$ In other words, if $a,b,c$ are non-negative real numbers, then how to prove ...
6
votes
2answers
619 views

Prove the inequality $a^2bc+b^2cd+c^2da+d^2ab \leq 4$ with $a+b+c+d=4$

Let $a,b,c$ and $d$ be positive real numbers such that $a+b+c+d=4.$ Prove the inequality $$a^2bc+b^2cd+c^2da+d^2ab \leq 4 .$$ Thanks :)
4
votes
1answer
117 views

$\frac {x_1^3}{y_1}+\frac {x_2^3}{y_3}+\ldots+\frac {x_n^3}{y_n}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x_1^2+x_2^2+\ldots+x_n^2) $

Let $b>a>0$ and $x_1, x_2,\ldots,x_n,y_1, y_2,\ldots,y_n\in [a,b]$. If $$x_1^2+x_2^2+\ldots+x_n^2=y_1^2+y_2^2+\ldots+y_n^2\,,$$ then $$\frac {x_1^3}{y_1}+\frac {x_2^3}{y_3}+\ldots+\frac {...
2
votes
3answers
146 views

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 ...
0
votes
3answers
387 views

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 ...
3
votes
4answers
1k views

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 ...