I have studied the following theorems on inequality: the QAGH (quadratic-arithmetic-geometric-harmonic) mean inequality, the triangle inequality, Cauchy Schwarz inequality, Weierstrass inequality and Chebycheff's sum inequality which states that  If $a_1,a_2,\ldots a_n$ are real numbers such that (i) $$a_1\le a_2\le\ldots \le a_n$$ and $$b_1\le b_2 \le \ldots \le b_n$$, then: $$n(a_1b_1 + a_2b_2 +\ldots + a_nb_n) \ge (a_1 + a_2 + \cdots + a_n)$$  (ii) $$a_1\ge a_2\ge\ldots\ge a_n$$ and $$b_1\le b_2 \le \ldots \le b_n$$, then: $$n(a_1b_1 + a_2b_2 +\ldots + a_nb_n) \le (a_1 + a_2 + \cdots + a_n)$$  One question from my assignment for which I need help is: If $$0\le x_1\le x_2\le\ldots\le x_n,$$ and it is given that $$\frac{1}{1+x_1} +\frac{1}{1+x_2}+\ldots+\frac{1}{1+x_n} = 1$$ Prove that: $$ (\sqrt{x_1} + \sqrt{x_2} +\cdots + \sqrt{x_n}) \ge (n - 1) \left(\frac{1}{\sqrt {x_1} } +\frac{1}{\sqrt{x_2} } + \cdots + \frac{1}{\sqrt{x_n} }\right) $$ Where n is a natural number, $ n\ge 2$ I have tried in many ways to prove this statement using the theorems I know, without any success. Could you please help me?
48460
Let $x_1,x_2,\ldots,x_n &gt; 0$ such that $\dfrac{1}{1+x_1}+\cdots+\dfrac{1}{1+x_n}=1$. Prove the following inequality. $$\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}+\cdots+\dfrac{1}{\sqrt{x_n}} \right ).$$  This is Exercice 1.48 in the book &quot;Inequalities-A mathematical Olympiad approach&quot;. Attempt I tried using HM-GM and I got $\left ( \dfrac{1}{x_1x_2\cdots x_n}\right)^{\frac{1}{2n}} \geq \dfrac{n}{\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}+\cdots+\dfrac{1}{\sqrt{x_n}}} \implies \dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_1}}+\cdots+\dfrac{1}{\sqrt{x_n}} \geq n(x_1x_2 \cdots x_n)^{\frac{1}{2n}}$. But I get stuck here and don't know if this even helps.
45099
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&gt;0$   My instinctive plan of attack is to use the AM/GM inequality with $x_1=\dfrac{a}{b+c}$ etc.  Using that I get this  $\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}}$   From here, I used the fact that $(a+b)(b+c)(a+c)\geq 8abc$, which can be easily proven by considering that $a+b\geq 2\sqrt{ab}$   But by using this, I get the following...  $\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}} \leq 3 \times \sqrt[3]{\dfrac{abc}{8abc}} = \dfrac{3}{2}$   Everything seems so perfect because I get the value $\dfrac{3}{2}$ as required, but this method isn't valid due to the change in direction! What is going on?  Is there a way of proving this inequality otherwise then?
46420
$(a+b+c+d+e+f)^2=$ sum of square of each number (X)+ $2($ sum of product of two numbers (Y) $)$ $ab+bc+cd+de+ef \le Y$ since all are positive.  Therefore $1\ge X+(ab+bc+cd+de+ef)$  Edit: From AM GM inequality, $$X\ge 6(abcde)^{1/3}$$and for am-gm inequality for a,b,c,d,e,f: $$1\ge 6(abcdef)^{1/6}$$ Hence $X\ge 6.\frac{1}{36}$ and $X\ge \frac{1}{6}$.   Is this correct?
40499
If $a, b, c$ belong to positive real numbers then prove that  $$\frac{2}{b+c} +\frac{2}{c+a}+\frac{2}{a+b} \le \frac 1a +\frac 1b +\frac 1c.$$  I applied AM-GM inequalities on both sides but I cant even figure out what to do on left side.
14394
Prove that  $(x_1+\dots+x_n)^2 \leq n(x_1^2 + \dots+x_n^2)$  for all positive integers n and all real numbers $x_1,....,x_n$  I am attempting a proof by induction but wasn't sure if i need the Cauchy-Schwarz Inequality or perhaps another way other than induction to prove this.  Proof   $n=1$ true  assume true for $n=k$  Now for $n= k+1$  $(x_1 + \dots +x_k + x_{k+1})^2 \leq \dots$
32334
This is supposed to be an application of AM-GM inequality.     if $abc=1$, then the following holds true: $a^2+b^2+c^2\ge a+b+c$   First of all,  $a^2+b^2+c^2\ge 3$  by a direct application of AM-GM.Also,we have  $a^2+b^2+c^2\ge ab+bc+ca$  Next,we consider the expression  $(a+1)(b+1)(c+1)=a+b+c+ab+bc+ca+abc\le a+b+c+a^2+b^2+c^2+1$  but that hardly helps.I know that  $3(a^2+b^2+c^2)\ge (a+b+c)^2$  From the first derived inequality,we know that the left hand side of the above inequality is greater than or equal to 9.But I can't see how that can be used here.I know that we can get $a+b+c\ge 3$ using AM-GM but that does not take me a step closer to finding the solution(from what I can understand).Also,  $a^3+b^3+c^3\ge a^2b+b^2c+c^2a$  $(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]\ge a^2b+b^2c+c^2a$ A hint will be appreciated at this point.
26784
Let $a, b$ and $c$ be posetive real numbers such that $abc=1$. I want to prove this inquality $$\sum_{cyc}\frac{ab}{a^5+b^5+ab}\le 1$$ I used AGM inequality but I didn't prove it.   Which should I use inequality? So thanks
48807
Let $a, b, c, d$ be positive real numbers such that $a^2 + b^2 + c^2 + d^2 = 4$, show that $$\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \ge 4.$$  My try:  $$\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \ge a + b + c + d,$$  yet $$a + b + c + d \le \sqrt{4(a^2 + b^2 + c^2 + d^2)} = 4.$$  Thus, direct application of Cauchy-Schwarz inequality is too weak. I tried other methods but with no significant progress:  $$(\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a)^2 \ge \frac {(a^{4/3} + b^{4/3} + c^{4/3})^3}{a^2 + b^2 + c^2 + d^2} = \frac {(a^{4/3} + b^{4/3} + c^{4/3})^3}4.$$  I also observed that  $$(\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a) + (\frac {a^2}c + \frac {b^2}d + \frac {c^2}a + \frac {d^2}b) + (\frac {a^2}d + \frac {b^2}a + \frac {c^2}b + \frac {d^2}c) + (\frac {a^2}a + \frac {b^2}b + \frac {c^2}c + \frac {d^2}d) = 4 (\frac 1a + \frac 1b + \frac 1c + \frac 1d) \ge 16,$$  since $$\frac 1a + \frac 1b + \frac 1c + \frac 1d \ge \sqrt{\frac {(1 + 1 + 1 + 1)^3}{a^2 + b^2 + c^2 + d^2}} = 4.$$  Now my work might seem stupid or off-topic here, but I provide it here because I wish any of these attempts will lead to a solution. Any hints will be appreciated.
79114
Let $n \ge3$ be an integer, and let $a_{2},a_{3}, ... ,a_{n}$ be positive real numbers such that $a_{2} a_{3}\cdots a_{n}=1.$ Prove that:   $$(1+a_{2})^{2}(1+a_{3})^{3}\cdots(1+a_{n})^{n}\ge n^n$$   This is the 2nd problem of the 53rd IMO and seems pretty interesting. How would we solve that?
116810
How to show  $x_1,x_2, \dots ,x_n \geq 0 $ and $ x_1 + x_2 + \dots + x_n \leq \frac{1}{2} \implies (1-x_1)(1-x_2) \cdots (1-x_n) \geq \frac{1}{2}$
11077
Prove the inequalities $|z_1 +z_2| \geq \frac{1}{2}(|z_1|+|z_2|)|\frac {z_1} {|z_1|} +\frac {z_2} {|z_2|}|$  Here is what I got so far  I start with $z_1=r_1 cis \theta_1$ (at this time we are not allowed to use $e^{i\theta}$ but $cis \theta =e^{i \theta}$)  $z_2 =r_2 cis \theta_2$. I will focus on the right hand side of the inequality   $RHS= \frac{1}{2}(r_1 +r_2)| \frac {r_1 cis \theta_1}{r_1} +\frac{r_2 cis \theta_2}{r_2}|=\frac{1}{2}(r_1 +r_2)|cis \theta_1 +cis \theta_2|=\frac{1}{2}(r_1 +r_2)((\cos \theta_1 +\cos \theta_2)^2 +(\sin \theta_1+\sin \theta_2)^2 )^{1/2}=\frac{1}{2}(r_1 +r_2)(2+2(cos \theta_1 cos \theta_2+ sin \theta_1 sin\theta_2))^{1/2}=\frac{1}{2}(r_1 +r_2)(2+2(cos (\theta_1 -\theta_2))^{1/2}=(r_1+r_2)(1+\cos(\theta_1-\theta_2))$  but I'm not sure how this result can help me prove the inequality, because I can't see how the left hand side related to this.
39982
I have an inequality problem which is as follow:  If $a,b,c$ are positive integers, with $a^2+b^2-ab=c^2$ prove that $(a-b)(b-c)\le0$.  I am not so good in inequalities. So, please give me some hints so that I can proceed.  Thanks.
6591
Please give me the proof for the following: Let $a_1,\,a_2,\,\dots\,a_n$ be $n$ positive real numbers whose product is equal to $1$. Prove that $$a_1+\cdots+a_n \geq n$$ and that the equality sign holds only if every $a_k$ is  equal to $1$.
76016
We know that for all $\varepsilon &gt; 0$, there exists an $N \in \mathbb{N}$ such that $\lvert x_n - 2 \rvert &lt; \epsilon$ for all $n \geq N$, and we want to show that for all $\varepsilon' &gt; 0$, there exists an $N' \in \mathbb{N}$ such that $\left| \frac{1}{x_n} - \frac{1}{2} \right| &lt; \epsilon'$ for all $n \geq N'$.  Let $\varepsilon = \varepsilon' - \frac{3}{2}$. Then by the triangle inequality, we have $$\left| \frac{1}{x_n} - \frac{1}{2} \right| \leq \left| \frac{1}{x_n}\right| + \left| -\frac{1}{2}\right|.$$  Because we proved earlier that $x_n &gt; 1$ for all $n \geq N$, we know that $$\left| \frac{1}{x_n}\right| + \left| -\frac{1}{2}\right| &lt; \left| x_n\right| + \left| -\frac{1}{2}\right|.$$  By the triangle inequality again, we know that $$\left| x_n - \frac{1}{2}\right| \leq \left| x_n\right| + \left| -\frac{1}{2}\right|.$$  Note that $\left| x_n - \frac{1}{2}\right| = \left| (x_n - 2) + (2 - \frac{1}{2})\right|$, so we get $$\left| (x_n - 2) + (2 - \frac{1}{2})\right| \leq \left| x_n - 2\right| + \left| 2 - \frac{1}{2}\right|.$$  Then we have $$\left| x_n - 2\right| + \left|2 - \frac{1}{2}\right| &lt; \varepsilon + \frac{3}{2} = \left(\varepsilon' - \frac{3}{2}\right) + \frac{3}{2} = \varepsilon',$$ so $\frac{1}{x_n} \rightarrow \frac{1}{2}$.  I just realized that I don't know how to make sure that $\varepsilon' - \frac{3}{2}$ is a positive number. How can I do this?
116601
If $a$ and $b$ are non-negative real numbers then show that $$(3a +\frac{4}{a+1}+\frac{8}{\sqrt{2(1+b^2)}})\cdot(3b +\frac{4}{b+1}+\frac{8}{\sqrt{2(1+a^2)}})\geq81.$$ Inequality is less common. To demonstrate the usual instruments (means inequality, ...) did not help. Does anyone have any idea?
92057
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 :)
22888
Prove that in any triangle $ABC$ the following inequality holds $$\cos{A}+\cos{B}+\cos{C}\ge\dfrac{1}{4}(3+\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})$$  And I have gotten $$8(\cos{A}+\cos{B}+\cos{C})\ge 6+2(\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})$$  $$2(\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})+3=(\sum_{cyc}\cos{A})^2+(\sum_{cyc}\sin{A})^2$$ $$\Longleftrightarrow 8\sum_{cyc}\cos{A}\ge 3+(\sum_{cyc}\cos{A})^2+(\sum_{cyc}\sin{A})^2$$ then Any hints, ideas? Thanks in advance.