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Questions tagged [rearrangement-inequality]

Proofs of inequalities by using Rearrangement inequality or Chebyshov inequality.

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21 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 ...
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1answer
34 views

$|\underset{i,j}{\sum}a_{ij}x_iy_j| \leq \underset{u,v \in \{-1,1\}^n }{sup} |\underset{i,j}{\sum} a_{ij}u_iv_j|$?

let $(a)_{ij}$ be a $M\times N$ Matrix with real entries ,is that possible to prove that: for any $x \in [-1,1]^n, y \in [-1,1]^m$ we have: $$|\underset{i,j}{\sum}a_{ij}x_iy_j| \leq \underset{u,v ...
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1answer
86 views

If $a+b+c=1$ and a,b,c >0 prove $\dfrac{b^2}{a+b^2}+\dfrac{c^2}{b+c^2}+\dfrac{a^2}{c+a^2} \geqslant \dfrac{3}{4}$ [duplicate]

If $a+b+c=1$ and a,b,c>0 prove $\dfrac{b^2}{a+b^2}+\dfrac{c^2}{b+c^2}+\dfrac{a^2}{c+a^2} \geqslant \dfrac{3}{4}$. I tried with CS Engel form,homogenization but ina anyway i can't prove inequality. Can ...
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2answers
79 views

Prove that $\sum_{cyc}\dfrac{a}{a + b^4 + c^4} \le 1$ where $abc = 1$.

If $a$, $b$ anc $c$ are three positives such that $abc = 1$ then prove that $$\large \sum_{cyc}\dfrac{a}{a + b^4 + c^4} \le 1$$ Here's what I did. $$\large \sum_{cyc}\dfrac{a}{a + b^4 + c^4}$$ $$\...
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2answers
89 views

Prove that $a\sqrt{b^3+1}+b\sqrt{c^3+1}+c\sqrt{a^3+1} \leq 5$

Let $a,b,c$ be nonnegative real numbers such that $a+b+c=3$. Prove that $$a\sqrt{b^3+1}+b\sqrt{c^3+1}+c\sqrt{a^3+1} \leq 5$$ I found a point at which the equality is attended, say $a=0,b=1,c=2$. But I ...
1
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1answer
126 views

Prove $\sum \sqrt{\frac{a^2}{6a^2+5ab+b^2}}\le \frac{\sqrt{3}}{2}$

Let $a,b,c\in R^+$ prove that the inequality $$\sqrt{\frac{a^2}{6a^2+5ab+b^2}}+\sqrt{\frac{b^2}{6b^2+5bc+c^2}}+\sqrt{\frac{c^2}{6c^2+5ca+a^2}}\le \frac{\sqrt{3}}{2}$$ My try:$$\sum\limits_{cyc} \sqrt{...
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2answers
50 views

Prove, that for every real numbers $ x \ge y \ge z > 0 $, and $x+y+z=\frac{9}{2}, xyz=1$, the following inequality takes place

Prove, that for every real numbers $ x \ge y \ge z > 0 $, and $x+y+z=\frac{9}{2}, xyz=1$, the following inequality takes place: $$ \frac{x}{y^3(1+y^2x)} + \frac{y}{z^3(1+z^2y) } + \frac{z}{x^3(1+...
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3answers
40 views

Sum of cross terms vs sum of squares? [closed]

What do we know about sum of squares vs sum of cross terms? Does one always dominate the other? Any theorems on that? e.g for $a^2 + b^2 + c^2 \ < ? > \ ab + ac + bc $ for any number of ...
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1answer
52 views

Find minimum of sum of product of sequences

Let $a_{i}, b_{i}, c_{i},\ d_{i}$ be non-negative sequences of length $k$ such that $$ \begin{matrix} \sum_{k}a_{i} & = & nk \\ \sum_{k}b_{i} & = & nk\\ \sum_{k}c_{i} & = &...
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2answers
73 views

If $a$, $b$ and $c$ are sides of a triangle, then prove that $a^\text{2}(b+c-a) + b^\text{2}(c+a-b) + c^\text{2}(a+b-c)$ $\leqslant$ $3abc$

Let $a$, $b$ and $c$ be the sides of a triangle. Prove that $$a^\text{2}(b+c-a) + b^\text{2}(c+a-b) + c^\text{2}(a+b-c) \leqslant 3abc$$ SOURCE: BANGLADESH MATH OLYMPIAD (Preparatory Question.) I am ...
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1answer
66 views

Inequality $\frac{a_1}{1^2}+\frac{a_2}{2^2}+…+\frac{a_n}{n^2}\ge\frac{1}{1}+\frac{1}{2}+…+\frac{1}{n}$ [duplicate]

Suppose $a_i$ are dinstinct positive integers $\forall1\le i\le n$. Prove that $$\frac{a_1}{1^2}+\frac{a_2}{2^2}+...+\frac{a_n}{n^2}\ge\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}$$ My approach: I ...
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2answers
64 views

Chebyshev's Sum Inequality Proof

So I was reading up on Chebyshev's Sum Inequality, and I was a little confused about the first proof presented on Wikipedia. Specifically, the line which reads "opening the brackets". What does this ...
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1answer
148 views

Prove that $\frac{a}{c\sqrt{a^2+1}}+\frac{b}{a\sqrt{b^2+1}}+\frac{c}{b\sqrt{c^2+1}}\ge \frac{3}{2}$

Let $a,b,c\in \Bbb R^+$ such that $a+b+c=abc$. Prove that $$\frac{a}{c\sqrt{a^2+1}}+\frac{b}{a\sqrt{b^2+1}}+\frac{c}{b\sqrt{c^2+1}}\ge \frac{3}{2}$$ Idea 1.From $a+b+c=abc\Leftrightarrow \frac{1}{ab}+...
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2answers
62 views

Algebraic inequality for positive reals $a,b,c$

The problem is from a previous maths olympiad and the last step is to prove the inequality $$4a^4bc + a^4c^2 + 9a^3bc^2 + 4a^3b^3 + 9a^2b^3c + a^2b^4 + 9ab^2c^3 + 4ab^4c + 4b^3c^3 + b^2c^4 +4a^3c^3 ...
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1answer
94 views

chebyshev's inequality - Question

I had a question in my exam and they asked to prove that prove that: $$3(1+a^2+a^4)\geq(1+a+a^2)^2$$ for all $a\in\mathbb R$. Now , I solved it , but the problem is that in the answer they wrote ...
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0answers
31 views

symmetrized rearrangement on sphere.

I am trying to undestand the Corollary 2.2 from Osgood, Phillips and Sarnak (see http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.486.558&rep=rep1&type=pdf), that is, if $u \in W^{1}(S^...
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1answer
49 views

prove the statement [closed]

Statement:If $a_1,a_2,a_3\cdots a_n$ be $n$ unequal and positive quantities and if $m>r>0$ , then $$\frac{a_1^{m}+a_2^{m}\cdots +a_n^{m}}{n}> \frac{a_1^{r}+a_2^{r}\cdots +a_n^{r}}{n}. \frac{...
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1answer
61 views

Prove $\sum \frac{a}{a+b^4+c^4} \le 1$

If $a,b,c \in \mathbb{R+}$ and $abc=1$ Prove That $$S=\sum \frac{a}{a+b^4+c^4} \le 1$$ My approach: we have $$S=\sum \frac{\frac{1}{bc}}{\frac{1}{bc}+b^4+c^4}=\sum \frac{1}{1+b^5c+bc^5}$$ Now by $...
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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 {...
1
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1answer
107 views

$\frac {x^3}{y}+\frac {z^3}{t}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x^2+z^2). $ [closed]

Let $0 < a < x, y, z, t < b $ s.t. $x^2+z^2= y^2+t^2$. Show that $$\frac {x^3}{y}+\frac {z^3}{t}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x^2+z^2). $$ I tried to apply Cauchy Schwartz. Also I ...
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2answers
268 views

Prove the inequality using Chebyshev's Inequality

If $a,b,c \in(0,\infty)$, then prove that: $$9(a^3+b^3+c^3)\ge(a+b+c)^3$$ I was trying to prove this inequality using Chebyshev's Inequality and assuming $a\ge b \ge c$ but to no avail. Can please ...
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1answer
73 views

How does the number $3$ help in this (probably Cauchy-based) inequality?

Given $a,b,c>0$ and $a+b+c=3$. Prove that $$\dfrac{a^2+bc}{b+ca}+\dfrac{b^2+ca}{c+ab}+\dfrac{c^2+ab}{a+bc}\ge3$$ Attempt: Using Cauchy inequality $(a+b\ge2\sqrt{ab})$: $\dfrac{a^2}{b+ca}+b+ca\...
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0answers
36 views

Optimization of sum of squares over permutations

Suppose I have fixed, positive values $n_1, \cdots, n_\ell$ and $T$. I'm looking for an algorithm to optimize \begin{align*} f(\boldsymbol{n}) = T\left(\sum_{j=1}^{\ell}\left(\sum_{i=1}^{j}n_i\right)^...
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2answers
61 views

Application of Chebyschev inequality

I want to prove the inequality above. On the extreme ends we get a clear application of AM-GM, and I want to use the chebyschev inequality for the middle but was having trouble. My Attempt: Since ...
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1answer
105 views

Prove $3\leqq x\sqrt{1+ y^{3}}+ y\sqrt{1+ z^{3}}+ z\sqrt{1+ x^{3}}\leqq 5$

This is an old problem of Pham Kim Hung! Prove: $$3\leqq x\sqrt{1+ y^{3}}+ y\sqrt{1+ z^{3}}+ z\sqrt{1+ x^{3}}\leqq 5$$ with $x,\,y,\,z\geqq 0$ & $x+y+z=3$ For the LHS, we have: $$\left \{ \sum\...
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3answers
127 views

Show that $a^2(a+b)^3+b^2(b+c)^3+c^2(c+a)^3\geq{\frac{8abc(a+b+c)^2}{3}}$ holds for $a,b,c>0$ [closed]

Let $a,b,c>0$. Show that $$a^2(a+b)^3+b^2(b+c)^3+c^2(c+a)^3\geq{\frac{8abc(a+b+c)^2}{3}}.$$ I have no idea. Please give a hint. Thanks!
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5answers
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Rearrangement inequality and minimal value of $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x}$

For $x \in \left(0, \dfrac{\pi}{2}\right)$, is the minimum value of $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x} = 1$? So considering ($\dfrac{1}{\cos x}$, $\dfrac{1}{\sin x}$) and ($\sin^3x$, $\...
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4answers
87 views

replacing Inequalities

I encountered a problem today: Prove that: $$\frac{a^3+b^3+c^3}{a^2+b^2+c^2} \ge \frac{a+b+c}{3}$$ for all $a,b,c>0$ I used the RMS-AM inequality to replace the LHS with $$\frac{\...
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5answers
50 views

Show that problem and that equality if fulfilled iff a=b

If $a, b >0$, show that $$ \frac{a}{b^2} + \frac{b}{a^2} \geq \frac{1}{a} +\frac{1}{b},$$ and that equality is fulfilled if and only if $a = b.$ I tried using elementary consequences of order ...
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1answer
39 views

Prove that $\sum_{cyc} {\frac{y-x}{y^2-1}}$ >0

If $x,y,z$ are real numbers, each greater than 1, then show that $\frac{y-x}{y^2-1}$+$\frac{z-y}{z^2-1}$+$\frac{x-z}{x^2-1}\gt 0$ It is not the actual problem,I deducted the actual problem in those ...
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3answers
116 views

How do I prove this inequality: $a^2+b^2+c^2 \geq ab+bc+ac$? [duplicate]

How do I prove that for any $a, b, c \in \mathbb{R}$ the inequality $a^2+b^2+c^2 \geq ab+bc+ac$ is true?
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2answers
109 views

Is it true that $\frac {a}{a+b}+\frac {b}{b+c}+\frac {c}{c+a}\geq \frac32$ if $a+b+c=1$?

Is it possible to prove that $ \dfrac {a}{a+b}+\dfrac {b}{b+c}+\dfrac {c}{c+a}\geq \dfrac {3}{2}$, if $ a+b+c=1, a,b,c>0$?
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0answers
21 views

Generalization of the Pólya–Szegő inequality

Let $d \in \mathbb{N}, 1\leq p \leq \infty$, $f \in C^{\infty}_0 (\mathbb{R}^d)$. It is known that there is a function $g$ such that $g(x)=g_0(|x|)$ for some non-increase function $g_0:[0, \infty) ...
4
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0answers
68 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 $n \leq 8$?

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 $n \le 8$? It is ...
1
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1answer
99 views

How is symmetry in an inequality determined?

I was reading a book about inequalities, in that I found that $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}>1\tag{1}$$ is a symmetric inequality in $a$,$b$,$c$. But if I change the order from $(a,b,c)$ to $...
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1answer
199 views

How to prove this interesting inequality: $\frac{5x+3y+z}{5z+3y+x}+\frac{5y+3z+x}{5x+3z+y}+\frac{5z+3x+y}{5y+3x+z}\ge 3$?

Here is my question: Prove that if $x,y,z>0$ then $$\frac{5x+3y+z}{5z+3y+x}+\frac{5y+3z+x}{5x+3z+y}+\frac{5z+3x+y}{5y+3x+z}\ge 3$$ Here is my attempt: Let us assume $x>y>z$: \begin{...
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
3answers
122 views

Find the minimum of expression: $\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z}$

If $x+y+z=1$ and $x,y,z$ are positive numbers, Find the minimum of expression: $$\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z}$$ My solution: $$\left[\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+...
0
votes
2answers
77 views

$\Big(\sum\limits_{i=1}^n \frac{1}{x_{i}^2} \Big) \Big(\sum\limits_{i=1}^n \frac{1}{x_{i}^{1/2}}\Big)^{2}$

Is there a way to combine the two following sums into only one sum? Perhaps an upper bound? $$A=\Big(\sum\limits_{i=1}^n \frac{1}{x_{i}^2} \Big) \Big(\sum\limits_{i=1}^n \frac{1}{x_{i}^{1/2}}\Big)^{2}...
1
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1answer
54 views

When does $\frac{1}{2n}\left(\sum_{i=1}^{n}x^2_i + \sum_{i=1}^{n}x_ix_{n-i+1}\right) \leq \left(\frac{1}{n}\sum_{i=1}^{n}x_i\right)^2$?

From brilliant.org, the Chebyshev's inequality, a generalization of the rearrangement inequality, states that for every nonincreasing sequences $\{a_i\}_{i=1}^n$ and $\{b_i\}_{i=1}^n$, $$ \frac{1}{n}...
3
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3answers
392 views

Proving that $\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}> \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

Prove that $\dfrac{ab}{c^3}+\dfrac{bc}{a^3}+\dfrac{ca}{b^3}> \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$, where $a,b,c$ are different positive real numbers. First, I tried to simplify the proof ...
4
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2answers
127 views

Prove $\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\leq\frac{x-1}{y-1}+\frac{y-1}{z-1}+\frac{z-1}{x-1}$

Let $x,y,z$ be real numbers all greater than $1$, then prove that $$\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\leq\frac{x-1}{y-1}+\frac{y-1}{z-1}+\frac{z-1}{x-1}$$ My Attempt: I am trying to ...
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1answer
63 views

prove the inequality for positive a, b, c

I faced with problem proving this inequality for positive $a$, $b$, $c$: $a\frac{a+b}{a+c} + b\frac{b+c}{a+b} + c\frac{c+a}{b+c} \geq a+b+c$ I tried to simplify it and I got that: $bc^3 + a^3 c + a ...
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 ...
0
votes
1answer
32 views

Smallest Sum of products, under cyclic permutations

Let $a_i$ be real positive numbers, $i=0, \dots, N-1$, with $N$ even. Define sums $S_k = \sum_{i=0}^{N-1} a_i a_{i+k}$, for $k=0, \dots, N-1$, where the indices are understood to be cyclic, i.e. $...
0
votes
3answers
384 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 ...
1
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3answers
72 views

Prove that: $\sum_{i=1}^n a_i^2\geq \sum_{i=1}^n a_ib_i$

Suppose that $a_1,a_2,...,a_n$ are positive real numbers.Also assume that $b_1,b_2,...b_n$ are an arbitrary permutation of $a_i$'s. Prove that: $$\sum_{i=1}^n a_i^2\geq \sum_{i=1}^n a_ib_i.$$ We ...
0
votes
2answers
122 views

How prove this inequality $x^3y+y^3z+z^3x\ge xyz(x+y+z)$

Let $x>0$, $y>0$ and $z>0$. Show that $$x^3y+y^3z+z^3x\ge xyz(x+y+z).$$ I known we can't WLOG: $x\ge y\ge z$, if this, I can use rearrangement inequality, But other I can't it. Thanks?
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2answers
129 views

If a, b, c>0 show that $\frac{a^2}{b^2+bc}+\frac{b^2}{c^2+ac}+\frac{c^2}{a^2+ab} \ge \frac{3}{2}$

For positive real numbers $a$, $b$, and $c$ prove that: $$\frac{a^2}{b^2+bc}+\frac{b^2}{c^2+ac}+\frac{c^2}{a^2+ab} \ge \frac{3}{2}.$$ I let $x=\frac{a}{b}$, $y=\frac{b}{c}$, and $z=\frac{c}{a}$. Then ...
3
votes
1answer
226 views

Inequality : $ (a_1a_2+a_2a_3+\ldots+a_na_1)\left(\frac{a_1}{a^2_2+a_2}+\frac{a_2}{a^2_3+a_3}+ \ldots+\frac{a_n}{a^2_1+a_1}\right)\geq \frac{n}{n+1} $

Let $a_1, a_2, \ldots, a_n $ be positive real numbers such that $\displaystyle\sum^n_{i=1} a_i=1$. Prove that $$ (a_1a_2+a_2a_3+\ldots+a_na_1)\left(\frac{a_1}{a^2_2+a_2}+\frac{a_2}{a^2_3+a_3}+ \...