All Questions
5
questions
1
vote
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68
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Proving $\sum_{i=1}^{n} \frac{a_{i}^2+a_{i+1}a_{i+2}}{a_{i}(a_{i+1}+a_{i+2})} \geq n.$
Given $a_1,a_2,...,a_n>0$, ($n\geq3, n \in \mathbb{N}$), prove that $$\frac{a_{1}^2+a_{2}a_{3}}{a_{1}(a_{2}+a_{3})}+\frac{a_{2}^2+a_{3}a_{4}}{a_{2}(a_{3}+a_{4})}+...+\frac{a_{n-1}^2+a_{n}a_{1}}{a_{...
- 235
3
votes
2
answers
178
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Prove that $\frac1{a(1+b)}+\frac1{b(1+c)}+\frac1{c(1+a)}\ge\frac3{1+abc}$
I tried doing it with CS-Engel to get $$
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geq \frac{9}{a+b+c+ a b+b c+a c}
$$
I thought that maybe proof that $$
\frac{1}{a+b+c+a b+b c+a c} \geq \...
3
votes
2
answers
81
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$AM-GM$-ish inequality
Suppose $x_0, \cdots, x_n$ are positive reals. Suppose that:
$$\sum_{i = 0}^n \frac{1}{1+x_i} \leq 1$$
Then show that:
$$\prod_{i=0}^{n} x_i \geq n^{n+1} $$
I got to this problem by rewriting problem ...
- 4,284
5
votes
3
answers
195
views
Prove that $\frac{\sqrt[n]{\prod_{k = 1}^nx_n}}{m} \ge n - 1$ where $\sum_{k = 1}^n\frac{1}{x_k + m} = \frac{1}{m}$.
Given positives $x_1, x_2, \cdots, x_{n - 1}, x_n$ such that $$\large \sum_{k = 1}^n\frac{1}{x_k + m} = \frac{1}{m}$$. Prove that $$\large \frac{\displaystyle \sqrt[n]{\prod_{k = 1}^nx_n}}{m} \ge n - ...
- 4,246
1
vote
1
answer
63
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Prove $x^{y+1}z+y^{z+1}x+z^{x+1}y\geq x^2y^2z^2$
Let $x>1$, $y>1$ and $z>1$ be such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$. Prove that:
$$x^{y+1}z+y^{z+1}x+z^{x+1}y\geq x^2y^2z^2.$$
When $x=y=z=3$, both sides are equal to $3^6$. ...
- 7,194