# Inequalities for idontknow, and for everyone else.

#### tahirimanov19

Mar 2015
182
68
Universe 2.71828i3.14159
1. If a, b, c are sides of the triangle, prove for any $x \in \mathbb{R}$
$$b^2x^2+(b^2+c^2-a^2)x+c^2>0$$

2. $$(a_1b_1+a_2b_2+a_3b_3)^2 \le ({a_1}^2+{a_2}^2+{a_3}^2)({b_1}^2+{b_2}^2+{b_3}^2)$$

3. Prove, if $a>0, \; b>0$,

$$\dfrac{a+b}{1+a+b} < \dfrac{a}{1+a} + \dfrac{b}{1+b}$$

4. If $a>b>0$ and $m>n$

$$\dfrac{a^m-b^m}{a^m+b^m} > \dfrac{a^n-b^n}{a^n+b^n}$$

5. $b_k>0, k=1,2,...,n$

$$min( \dfrac{a_1}{b_1} , \dfrac{a_2}{b_2} , ... , \dfrac{a_n}{b_n})<\dfrac{a_1+a_2+...+a_n}{b_1+b_2+...+b_n} < max( \dfrac{a_1}{b_1} , \dfrac{a_2}{b_2} , ... , \dfrac{a_n}{b_n})$$

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#### idontknow

Dec 2015
1,076
166
Earth
First about 3. $$\displaystyle \frac{a+b}{1+a+b}=\frac{a}{1+a+b}+\frac{b}{1+a+b}<\frac{a}{1+a}+\frac{b}{1+b}$$. (we can avoid a or b from denominator since it is greater than 1 for each a,b)

2. $$\displaystyle {\displaystyle \left|\sum _{i=1}^{n}u_{i}{\bar {v}}_{i}\right|^{2}\leq \sum _{j=1}^{n}|u_{j}|^{2}\sum _{k=1}^{n}|v_{k}|^{2}}\;$$ , now set $$\displaystyle n=3$$ and $$\displaystyle u_i = a_i$$ and $$\displaystyle v_i = b_i$$.

1.To save time(avoid trigonometry) it is enough to write the expression like $$\displaystyle b^2 \cdot (x-x_1 )(x-x_2 )>0$$ and see whether it holds true for any $$\displaystyle x\in R$$.

Maybe someone else can finish 5 and 4.

#### idontknow

Dec 2015
1,076
166
Earth
4. and 5. are left.

#### greg1313

Forum Staff
Oct 2008
8,008
1,174
London, Ontario, Canada - The Forest City
We don't mind these sorts of posts.

Last edited:
Reactions: 1 person
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