# Questions tagged [rearrangement-inequality]

Proofs of inequalities by using Rearrangement inequality or Chebyshov inequality.

92 questions
3k views

### If both $a,b>0$, then $a^ab^b \ge a^bb^a$ [closed]

Prove that $a^a \ b^b \ge a^b \ b^a$, if both $a$ and $b$ are positive.
369 views

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 ...
211 views

### Proving with AM-GM Inequality

$$\frac{4}{abcd}\geq\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}$$ Given: $a+b+c+d=4$ and $a$, $b$, $c$ abd $d$ are positives. How to prove the above inequality using Arithmetic Geometric Mean ...
316 views

160 views

99 views

### Chebyshev Inequality toughnut [closed]

Let $a^2 + b^2 + c^2 + d^2 = 1$, where $(a,b,c,d \geq 0)$. Prove that: $$\frac{a^{2}}{b+c+d}+\frac{b^{2}}{a+c+d}+\frac{c^{2}}{b+a+d}+\frac{d^2}{b+c+a} \geq \frac{2}{3}$$
175 views

### Prove that $A_1 \cdot B_1 + A_2 \cdot B_2 > A_1 \cdot B_2 + A_2 \cdot B_1$ for $0 < A_1 < A_2$ and $0 < B_1 < B_2$
We are asked to prove or disprove that this is correct $A_1 \cdot B_1 + A_2 \cdot B_2 > A_1 \cdot B_2 + A_2 \cdot B_1$ for $0 < A_1 < A_2$ and $0 < B_1 < B_2$. I'm not very ...