Questions tagged [rearrangement-inequality]

Proofs of inequalities by using Rearrangement inequality or Chebyshov inequality.

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If $a,b,c>0$ and $a+b+c=1$, then prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le 3+2\cdot\frac{a^3+b^3+c^3}{abc}$.

Question: If $a,b,c>0$ and $a+b+c=1$, then prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le 3+2\cdot\frac{a^3+b^3+c^3}{abc}$. Solution: Observe that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le 3+2\...
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2answers
39 views

Proving an inequality regarding increasing or decreasing sequences

Suppose $a_1 \geq a_2 $ and $b_1 \geq b_2$. Prove that $$ (a_1+a_2)(b_1+b_2) \leq 2 (a_1b_1 + a_2 b_2) $$ Generalize sol attempt: We can write $a_1 b_1 + a_2 b_2 + a_2 b_1 + a_1 b_2 \leq ...
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2answers
91 views

If $a+b+c=3$ Prove that $a^{2}+b^{2}+c^{2}\geq\frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}$

Question - Let $a, b, c$ be positive real numbers such that $a+b+c=3 .$ Prove that $$ a^{2}+b^{2}+c^{2} \geq \frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a} $$ My try - i tried putting $a+2 = x,...
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2answers
55 views

$ \frac{a^{2}}{a+2 b^{3}}+\frac{b^{2}}{b+2 c^{3}}+\frac{c^{2}}{c+2 a^{3}} \geq 1 $

Question Let.a, $b, c$ be positive real numbers with sum 3 . Prove that $$ \frac{a^{2}}{a+2 b^{3}}+\frac{b^{2}}{b+2 c^{3}}+\frac{c^{2}}{c+2 a^{3}} \geq 1 $$ my doubt - by using cauchy reverse ...
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1answer
31 views

Where is the flaw in this using Rearrangement Inequality?

I assume that the reader understands by Rearrangement Inequality that if $a_i$ and $b_i$ are reals, and $a_1 ≤ a_2 ≤ ... ≤ a_n $ and $b_1 ≤ b_2 ≤ ...≤ b_n$ then $\Sigma_{i=0}^{i=n} a_i × b_i$ ≥ $\...
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1answer
99 views

For $a,b,c\in\left[\frac{1}{\sqrt{6}}, 6\right]$: $\sum_{cyc}\frac{4}{a+3b}\geq \sum_{cyc}\frac{3}{a+2b}$

For $a,b,c\in\left[\frac{1}{\sqrt{6}}, 6\right]$ prove that $$\frac{4}{a+3b}+\frac{4}{b+3c}+\frac{4}{c+3a}\geq\frac{3}{a+2b}+\frac{3}{b+2c}+\frac{3}{c+2a}.$$ I can't really find a way to exploit the ...
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1answer
73 views

Inequality from Israel TST

Let $a, b, c, d$ be nonnegative numbers such that $a+b+c+d=18.$ Prove that: $$\sqrt{\frac{a}{b+6}}+\sqrt{\frac{b}{c+6}}+\sqrt{\frac{c}{d+6}}+\sqrt{\frac{d}{a+6}}\leq5\sqrt{\frac{2}{7}}$$ These are my ...
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1answer
85 views

One of my old inequality (very sharp)

I'm proud to present one of my old inequality that I can't solve : Let $a,b,c>0$ such that $a+b+c=1$ and $a\ge b \geq c $ then we have :$$\sqrt{\frac{a}{a^a+b^b}}+\sqrt{\frac{b}{b^b+c^c}}+\...
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1answer
32 views

How many Steiner Symmetrizations does it take to make an arbitrary set convex?

I have not seen this question investigated before but I might be wrong: Can any subset of $\mathbb{R}^d$ be turned into a convex set by finitely many steiner symmetrizations? If yes, is the number of ...
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2answers
42 views

Natural numbers inequality $na^{n-1}b\leq(n-1)a^n+b^n$ by induction

Let $a$ and $b$ be arbitrary natural numbers and $n$ some positive integer. How to prove the inequality $$na^{n-1}b\leq(n-1)a^n+b^n$$ by induction for all $n$? This is related to this result, and, ...
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2answers
76 views

$\frac{a}{a+2b+c}+\frac{b}{b+2c+a}+\frac{c}{c+2a+b}\geq\frac{3}{4}$

I found the following exercise: Prove that $$\frac{a}{a+2b+c}+\frac{b}{b+2c+a}+\frac{c}{c+2a+b}\geq\frac{3}{4}$$ for any positive $a$, $b$, $c$. I tried substituting the denominators but it led me ...
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Can I do this? Divide the terms inside $ (\frac {(m-a)^2}{2b} + d)^N $ by $d$?

For part $d)$ of this question I began with the equation in part c and divided all terms within the bracket by $d$. I then used the substitution $d = \frac {c}{E[r]}$ and arrived at the stated answer. ...
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2answers
46 views

Proving an inequality involving fractions and square roots holds

I have tried to prove the following inequality holds using a few approaches but none have worked. I am not really sure if I'm missing something. Here's the question: For every $x, y > 0$ prove ...
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23 views

Rearrangement inequality with infinite variables

Is the rearrangement inequality generalizable for infinite number of variables? In other words if I have an infinite sum $a_1x_1 + a_2x_2 + ... $ And we can change the order of the coefficients. ...
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1answer
81 views

Largest integer $k$ such that $\frac{a^{m+1}+b^{m+1}}{a^m+b^m}\geq\sqrt[k]{\frac{a^k+b^k}2}$

The setup is as follows: Suppose that $m$ is a given natural number. What is the greatest natural number $k$ such that for all real numbers $a,b>0$, we have $$\sqrt[k]{\frac{a^k+b^k}2}\le\frac{a^{...
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3answers
66 views

The inequality $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \frac{3}{2}$ [duplicate]

I was told that the following inequality $$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \frac{3}{2}$$ can be solved by the rearrangement inequality https://en.wikipedia.org/wiki/...
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An elegant but difficult inequality. [duplicate]

$$\frac{x}{\sqrt{x+y}} + \frac{y}{\sqrt{y+z}} + \frac{z}{\sqrt{z+x}}$$ Over all non-negative $x,y,z$ satisfying $x+y+z=4$, let the maximum value of the above expression be $M$. What is the value of $...
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2answers
214 views

Typical Olympiad Inequality? If $\sum_i^na_i=n$ with $a_i>0$, then $\sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n$

Let $\sum_i^na_i=n$, $a_i>0$. Then prove that $$ \sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n $$ I have tried AM-GM, Cauchy-Schwarz, Rearrangement etc. but nothing seems to work. The ...
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2answers
219 views

Inequality : $\frac{a}{\exp(a+b)}+\frac{b}{\exp(b+c)}+\frac{c}{\exp(c+a)}\leq \exp\Big(\frac{-2}{3}\Big)$

It's a charming problem : Let $a,b,c>0$ such that $a+b+c=1$ then we have : $$\frac{a}{\exp(a+b)}+\frac{b}{\exp(b+c)}+\frac{c}{\exp(c+a)}\leq \exp\Big(\frac{-2}{3}\Big)$$ I know the identity : ...
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3answers
204 views

Nice olympiad inequality :$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{3}{7}$

I have this to solve : Let $x,y,z>0$ such that $x+y+z=3$ then we have : $$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{3}{7}$$ I try to use Jensen's inequality but ...
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1answer
72 views

Maximizing a permutation of a quadratic form

Let \begin{align*} f_A(x) = x^\intercal A x \end{align*} for some positive semi-definite $A \in \mathbb{R}^{n\times n}$. Unlike the standard problem \begin{align*} f^* = \max_{x: \|x\| = 1}f_A(x) \end{...
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1answer
83 views

Inequalities: BMO 2009/10 Round 2 [duplicate]

BMO 2009/10 Round 2 Q.4 asks Prove that for all positive reals $x,y, z$: $$4(x+y+z)^3 >27(x^2y + y^2z + z^2x)$$ My try:- Using AM-GM, LHS is greater than $108xyz$. Using AM-GM, RHS is greater ...
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4answers
168 views

Calculate the maximum value of $x^3y + y^3z + z^3x$ where $x + y + z = 4$ and $x, y, z \ge 0$.

Given non-negatives $x, y, z$ such that $x + y + z = 4$. Calculate the maximum value of $$\large x^3y + y^3z + z^3x$$ As an assumption, the maximum value is $27$, occured when $(x, y, z) = (0, 1, 3)$....
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6answers
103 views

minimum value of $\frac{x}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}$ is [duplicate]

If $x,y,z>0$ and $x+y+z=1.$ Then minimum value of $\displaystyle \frac{x}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}$ is What i try From Titu lema $$\frac{x^2}{2x-x^2}+\frac{y^2}{2y-y^2}+\frac{z^2}...
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0answers
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Prove that $\sum_{cyc}\frac{m^2}{n + p^2} \ge \sum_{cyc}\frac{m}{n^2 + p}$ where $m, n, p \ge 0$ and $mnp = 1$.

Given positives $m, n, p$ such that $mnp = 1$, prove that $$\large \frac{m^2}{n + p^2} + \frac{n^2}{p + m^2} + \frac{p^2}{m + n^2} \ge \frac{m}{n^2 + p} + \frac{n}{p^2 + m} + \frac{p}{m^2 + n}$$ I ...
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2answers
133 views

Given three non-negative numbers $a, b, c$ so that $a+ b+ c= 3,\,a^{2}+ b^{2}+ c^{2}= 5$. Prove $a^{3}b+ b^{3}c+ c^{3}a\leqq 8$ .

Problem. Given three non-negative numbers $a, b, c$ so that $a+ b+ c= 3,\,a^{2}+ b^{2}+ c^{2}= 5$. Prove: $$a^{3}b+ b^{3}c+ c^{3}a\leqq 8$$ My solution in M&Y : (and I'm looking forward to seeing ...
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4answers
79 views

How to prove this inequality for $a,b,c>0$?

How to prove the inequality for $a,b,c>0$ : $$\frac{2a-b-c}{2(b+c)^2}+\frac{2b-a-c}{2(a+c)^2}+\frac{2c-b-a}{2(b+a)^2}\geq 0$$ ?
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4answers
93 views

Calculate the minimum value of $\sum_\mathrm{cyc}\frac{a^2}{b + c}$ where $a, b, c > 0$ and $\sum_\mathrm{cyc}\sqrt{a^2 + b^2} = 1$.

$a$, $b$ and $c$ are positives such that $\sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{c^2 + a^2} = 1$. Calculate the minimum value of $$\frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b}$$ I ...
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Is there a problem in this statement: As $\sum_{n=1}^{\infty} \mu(n)/n = 0, \sum_{n=1}^{N} \mu(n)/n = -\sum_{n=N+1}^{\infty} \mu(n)/n$.

As by Landau's proof $$\sum_{n=1}^{\infty} \mu(n)/n = 0$$ Therefore for any $N \in \mathbb{N}$, $$ \sum_{n=1}^{N} \mu(n)/n = -\sum_{n=N+1}^{\infty} \mu(n)/n$$ Is there a problem with the above ...
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1answer
126 views

Given three postive numbers $a,b,c$ so that $a\geqq b\geqq c$. Prove that $\sum\limits_{cyc}\frac{a+bW}{aW+b}\geqq 3$ . [closed]

Given three postive numbers $a, b, c$ so that $a\geqq b\geqq c$. Prove that $$\sum\limits_{cyc}\frac{a+ b\sqrt{\frac{b}{c}}}{a\sqrt{\frac{b}{c}}+ b}\geqq 3$$ I make it Firstly, we need to have one ...
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3answers
28 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 ...
1
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1answer
37 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
96 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 ...
2
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2answers
127 views

Prove that $\sum_{cyc}\frac{a}{a + b^4 + c^4} \le 1$ where $abc = 1$ [duplicate]

If $a$, $b$ anc $c$ are three positives such that $abc = 1$ then prove that $$ \sum_{cyc}\dfrac{a}{a + b^4 + c^4} \le 1$$ Here's what I did. $$ \sum_{cyc}\frac{a}{a + b^4 + c^4}$$ $$\le \sum_{cyc}\...
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2answers
135 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
159 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{...
1
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2answers
56 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+...
1
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3answers
67 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 ...
0
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1answer
62 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} & = &...
4
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2answers
173 views

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

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 ...
3
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1answer
67 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 ...
1
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2answers
219 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 ...
2
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1answer
171 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}+...
0
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2answers
67 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 ...
4
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1answer
105 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 ...
1
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0answers
46 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
57 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{...
0
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1answer
76 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 $...
4
votes
1answer
134 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
132 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 ...