Questions tagged [sum-of-squares-method]

Proofs of inequalities by the Sum of Squares method (SOS).

Filter by
Sorted by
Tagged with
2
votes
0answers
31 views

Rational rank one decomposition of symmetric positive semidefinite integer matrices

Problem: Given an $n\times n$ symmetric positive semidefinite (PSD) $\color{blue}{\textbf{integer}}$ matrix $Q$ with $\mathrm{Rank}(Q) = r$, find $\color{blue}{\textbf{integer}}$ vectors $u_i, i=1, \...
2
votes
0answers
20 views

SOSTOOLS - Express a Matrix polynomials as Polynomial

We know that to express an polynomial as SOS, we can write it as a Matrix SOS Now we take an example, use SOStools in Matlab to get SOS for $$f(a,b,c)=16(a^2+b^2+c^2)^3-9\left[(a^3+3b^2c)^2+(3ac^2+b^3)...
0
votes
1answer
28 views

Simplifying this equation to get sum of squares

$$\frac{1}{n\sum x_{i}^2-{(\sum x_{i}})^2}$$ I have this equation above. I am trying to simplify it such that I can get: $\dfrac{1}{nSS_{x}}$ Where $SS_{x}$ is the sum of squares of $x$. Any pointers ...
2
votes
2answers
96 views

Express $169$ as the sum of $1,2,3,4,5$ non-zero squares

I'm trying to solve the following exercise. Show that $169$ can be expressed as a sum of $1,2,3,4,5$ non-zero squares, and deduce that any $n \ge 169$ is the sum of five non-zero squares. The latter ...
1
vote
4answers
106 views

If$x^2+y^2+z^2+t^2=x(y+z+t)$prove $x=y=z=t=0$

If $$x^2+y^2+z^2+t^2=x(y+z+t)$$ Prove $x=y=z=t=0$ I added $x^2$ to both side of the equation: $$x^2+x^2+y^2+z^2+t^2=x(x+y+z+t)$$ Then rewrite it as: $$x^2+(x+y+z+t)^2-2(xy+xz+xt+yz+yt+zt)=x(x+y+z+t)$$...
27
votes
4answers
791 views

Proving $abcd+3\geq a+b+c+d$

If $a,b,c,d$ are non negatives and $a^2+b^2+c^2+d^2=3$ prove that $$abcd+3\ge a+b+c+d$$ The inequality is not as simple as it looks.The interesting part is that the equality occurs when $a=0,b=c=d=1$...
3
votes
4answers
210 views

Prove the polynomial $x^4+4 x^3+4 x^2-4 x+3$ is positive

Given the following polynomial $$ x^4+4 x^3+4 x^2-4 x+3 $$ I know it is positive, because I looked at the graphics and I found with the help of Mathematica that the following form $$ (x + a)^2 (x + b)...
0
votes
0answers
23 views

If the real zeroes of real polynomial p(x,y) are disjoint points and curves, is p(x,y) a positive sum of squares?

For example, $p(x,y) = x^2(x-1)^2 + y^2(y-1)^2$ has real zeroes in the set $\{(0,0), (0, 1), (1, 0), (1, 1)\}$ and admits a decomposition into a sum of squares. How can I find decompositions like ...
1
vote
1answer
84 views

Inequality $\frac{xy+z}{x+yz}+\frac{yz+x}{y+zx}+\frac{zx+y}{z+xy}-\frac{x+y+z}{3}\leq 1$

For $x,y,z \in [2,\infty)$, prove that $\frac{xy+z}{x+yz}+\frac{yz+x}{y+zx}+\frac{zx+y}{z+xy}-\frac{x+y+z}{3}\leq 1$ I tried to group the terms and prove that $\frac{xy+z}{x+yz} - \frac{y}{3}\leq \...
0
votes
2answers
113 views

prove thatt $\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$

prove that $$\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$$ where $x,y,z$ are positives such that $xy+yz+xz=1$ By Holder;$$\left(\sum_{cyc} \frac{1}{{(x+y)}^2} \right){\left(\sum yz+zx \right)}^2\ge {\sum \...
1
vote
5answers
82 views

$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$ Proof

Does anyone know hot to prove this inequality? Having: $a, b, c \gt 0$ $$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$$ I tried with the AM-GM inequality but I couldn't get ...
1
vote
1answer
94 views

Prove $3\left(9-5\sqrt{3}\right) \sum \frac{1}{a} \geqslant \sum a^2+\frac32\cdot\frac{\left[(\sqrt3-2)(ab+bc+ca)+abc\right]^2}{abc}$

Let $a,\,b,\,c$ are positive real numbers satisfy $a+b+c=3.$ Prove that $$3\left(9-5\sqrt{3}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant a^2+b^2+c^2 + \frac32 \cdot \frac{\left[(\...
3
votes
2answers
110 views

Prove $5\Big(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\Big)\geq \frac{a^2+b^2+c^2}{ab+bc+ca}+10.$

Problem. (?) For $a,b,c$ be non-negative numbers such as $a \geq 2(b+c).$ Prove:$$5\Big(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\Big)\geq \frac{a^2+b^2+c^2}{ab+bc+ca}+10.$$ My Solution. We write the ...
0
votes
3answers
45 views

Proving a cyclic inequality

Show that $a^4 + b^4 + c^4 \geq a^3b + b^3c + c^3a$ for any postive integers $a, b, c$ I'm not sure how to approach this problem. I've tried assuming that WLOG $a > b > c$ so that it is clear ...
2
votes
2answers
112 views

For $a,b,c>0$ proving $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geqslant a + b + c + \frac{4(a - b)^2}{a + b + c}$ [duplicate]

The problem with which I have a problem it's this: For $a,b,c>0$ prove that $$ \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geqslant a + b + c + \frac{4(a - b)^2}{a + b + c} $$ Titu's Lemma ...
1
vote
1answer
56 views

How to formulate Augmented Langrangian as an objective function to SOS problem?

I am implementing ADMM algorithm to solve a complex optimization problem with SOS constraints. The x-update of the algorithm looks like this: $$ x^{k+1}= {\arg\min}_{x \in S} \|x-z^{k}+\lambda^{k}\|^...
0
votes
2answers
76 views

Sum of two squares question

Which of the following statements is true for every value of $n$? A: If $n$ is not a sum of two squares, then neither is 69$n$ B: If $n$ is a sum of two squares, then so is 34$n$ C: If $n$ is not a ...
1
vote
0answers
43 views

Expressing a polynomial as a sum of squares in Maple

While sostools in MATLAB would find such a sum of squares decomposition, I am wondering whether a similar package exists for Maple. Example. Express the following polynomial as Sum Of Squares:$$\frac{...
3
votes
0answers
66 views

Prove $\frac{4}{a^2 + 2b^2 + 3c^2 + 10} \le \frac{5a + 3b + c + 7d}{16(a+b+c+d)}$ for positive reals $abcd=1$

Problem: Let $a, b, c, d > 0$ with $abcd = 1$. Prove that $$\frac{4}{a^2 + 2b^2 + 3c^2 + 10} \le \frac{5a + 3b + c + 7d}{16(a+b+c+d)}.$$ Background Information: It is verified by Mathematica. This ...
0
votes
2answers
62 views

Explanation to an unmotivated step in an inequality.

Follows the original problem: Let $a, b, c$ be non-negative real numbers. Prove that $$ \frac{a^2}{a^2 + 2\left(a + b\right)^2} + \frac{b^2}{b^2 + 2\left(b + c\right)^2} + \frac{c^2}{c^2 + 2\left(c + ...
3
votes
1answer
117 views

A cyclic inequality of degree 10

Suppose that $x,y,z\geq 0$. I would like to prove that $$(x^5+y^5+z^5)^2\geq (x+y+z)(x^3y^6+y^3z^6+z^3x^6).$$ I can prove this inequality using some standard methods. For example, I can let $x=1, y=1+...
4
votes
3answers
80 views

$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq \frac{3}{2}$ for $a,b,c\in\mathbb{R}^+$ with $abc=1$

Suppose that $a,b,c$ are positive reals such that $abc=1$. Prove that $$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq \frac{3}{2}.$$ Hint: Use Titu's lemma. My approach: I am trying to use Titu'...
0
votes
2answers
66 views

Symmetric inequality over 3 variables with restriction $a+b+c=1$

Suppose that $a,b,c$ are positive real numbers with $a+b+c=1$. Prove that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq 3+\frac{2(a^3+b^3+c^3)}{abc}.$$ Since I am texting from cellphone I will skip some ...
1
vote
1answer
54 views

question from Euclid 2011 about proving that $\frac{(a^3+b^3)c}{a^2+b^2}+\frac{(b^3+c^3)a}{b^2+c^2}+\frac{(c^3+a^3)}{c^2+a^2}\ge 1$

I just did the following question: If $a, b, c$ positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}$ prove that $\frac{(a^3+b^3)c}{a^2+b^2}+\frac{(b^3+c^3)a}{b^2+c^2}+\...
2
votes
4answers
95 views

To Prove $\frac{1}{b}+\frac{1}{c}+\frac{1}{a} > \sqrt{a}+\sqrt{b}+\sqrt{c}$

The three sides of a triangle are $a,b,c$, the area of the triangle is $0.25$, the radius of the circumcircle is $1$. Prove that $1/b+1/c+1/a > \sqrt{a}+\sqrt{b}+\sqrt{c}$. what I've tried: $$\frac{...
1
vote
0answers
105 views

Proving Markov–Lukacs theorem from SOS programming

Let $f(t)$ be a univariate polynomial of even degree $n$. Markov-Lukacs theorem (reference,Eq.45) says this: $f(t)$ is nonnegative in $t\in[-1,1]$ iff $$ f(t)=g(t)^2+(t+1)(1-t)h(t)^2 $$ where $g$ and $...
2
votes
4answers
98 views

Proving $\frac{{a}^3+{b}^3+ac(a-c)+bc(b-c)-5abc}{{(a+b+c)}^3} \geqslant -\frac{1}{5}$

For $a,b,c\geqslant 0.$ Prove$:$ $$\dfrac{{a}^3+{b}^3+ac(a-c)+bc(b-c)-5abc}{{(a+b+c)}^3} \geqslant -\dfrac{1}{5}$$ I found an AM-GM proof. Since $$P+\frac{1}{5}\geqslant 0\Leftrightarrow 6\,{a}^{3}+6\,...
0
votes
1answer
68 views

Proving $\sum \frac{a+b}{c} \geq 2.\sqrt{(a+b+c)(\frac{a}{bc} +\frac{b}{ca}+ \frac{c}{ab})}$

Problem. (Le Khanh Sy) For $a,b,c>0.$ Prove$:$ $$\sum \dfrac{a+b}{c} \geq 2\sqrt{(a+b+c)\Big(\dfrac{a}{bc} +\dfrac{b}{ca}+ \dfrac{c}{ab}\Big)}$$ My proof. After squaring ... it's $$4\,{b}^{2}{c}^{2}...
4
votes
3answers
93 views

Proving $(a+b+c) \Big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Big) \leqslant 25$

For $a,b,c \in \Big[\dfrac{1}{3},3\Big].$ Prove$:$ $$(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) \leqslant 25.$$ Assume $a\equiv \text{mid}\{a,b,c\},$ we have$:$ $$25-(a+b+c) \Big(\dfrac{...
5
votes
6answers
149 views

prove $a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b)$

prove $$a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b),$$$a,b,c>0$ Obviously this is a direct consequence of the third degree schur's inequality. I was wondering if this could be proved without this ...
2
votes
2answers
146 views

Prove $\sum ab \sum \frac{1}{(a+b)^2} \geqslant \frac{9}{4}+\frac{kabc\sum (a^2-bc)}{(a+b+c)^3(ab+bc+ca)}$ for the best k.

For $a,b,c\geqslant 0;ab+bc+ca>0.$ Find $k_\max$ and proving in that case$:$ $$(ab+bc+ca)\Big(\dfrac{1}{(a+b)^2}+\dfrac{1}{(b+c)^2}+\dfrac{1}{(c+a)^2}\Big) \geqslant \dfrac{9}{4}+\dfrac{kabc(a^2+b^...
2
votes
1answer
73 views

Proving $2\left(b^2+c^2\right)-a^2\leqslant 12$ with some condition.

Problem. Let $a,b,c\in\mathbb{R}$ such that $a+b+c=6,$ $a^2+b^2+c^2\in\left[12,\frac{68}3\right]$ and $a\geq b\geq c.$ Prove $$2\left(b^2+c^2\right)-a^2\leqslant 12.$$ When do we have equality? I can ...
4
votes
2answers
155 views

Proving $\frac{a}{b^3}+\frac{b}{c^3}+\frac{c}{a^3}\geqslant \frac{a+b}{b^3+c^3}+\frac{b+c}{c^3+a^3}+\frac{c+a}{a^3+b^3}$

For $a,b,c>0.$ Prove$:$ $$\dfrac{a}{b^3}+\dfrac{b}{c^3}+\dfrac{c}{a^3}\geqslant \dfrac{a+b}{b^3+c^3}+\dfrac{b+c}{c^3+a^3}+\dfrac{c+a}{a^3+b^3}\quad (\text{Tran Quoc Thinh}) $$ It's easy with ...
4
votes
2answers
114 views

Proving $4\Big(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \Big)+\frac{81}{(a+b+c)^2}\geqslant{\frac {7(a+b+c)}{abc}}$

For $a,b,c>0.$ Prove$:$ $$4\Big(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \Big)+\dfrac{81}{(a+b+c)^2}\geqslant{\dfrac {7(a+b+c)}{abc}}$$ My proof is using SOS$:$ $${c}^{2}{a}^{2} {b}^{2}\Big( \...
1
vote
5answers
115 views

SOS proof for $\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$

I need an SOS(sum of squares) proof for $$\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$$ if $a,b,c>0$ I already have a am-gm proof but is there a way to use SOS. Am-gm proof : $\frac{a^3}{bc}+b+c\ge 3a$ ...
3
votes
4answers
107 views

Is this alternative proof of the inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$ correct?

Prove that for all positive real numbers: $$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq\dfrac{3}{2}$$ This is same as this question but a different approach is used there whereas I want to verify ...
2
votes
2answers
95 views

Proving $Q=\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+k\cdot \frac{(ab+bc+ca)}{(a+b+c)^{2}}\geqslant \frac{3}{8}+\frac{k}{3}$

For $a,b,c\geqslant 0.$ Prove: $$\text{Q}=\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+k\cdot \frac{(ab+bc+ca)}{(a+b+c)^{2}}\geqslant \frac{3}{8}+\frac{k}{3},$$ where $k={\frac {27}{8}}+\frac{9\sqrt{3}}{...
3
votes
1answer
100 views

Proving ${\frac {35{x}^{2}+7x(y+z)+23yz}{35(x^2+y^2+z^2)+37(xy+yz+zx)}}\leqslant \sqrt {{\frac {{x}^{2}+yz}{6\,{y}^{2} +6\,yz+6\,{z}^{2}}}}$

For $x,y,z \geqslant 0.$ Proving$:$ $${\dfrac {35{x}^{2}+7x(y+z)+23yz}{35(x^2+y^2+z^2)+37(xy+yz+zx)}}\leqslant \sqrt {{\dfrac {{x}^{2}+yz}{6{y}^{2} +6yz+6{z}^{2}}}}\quad \quad(\text{tthnew})$$ My ...
3
votes
1answer
143 views

Proving $\sum \frac{b+c}{9(a^2+3bc)+4(a+b+c)^2}\geqslant \frac{1}{4(a+b+c)}$

For $a,b,c>0.$ Prove$:$ $$\sum \dfrac{b+c}{9(a^2+3bc)+4(a+b+c)^2}\geqslant \dfrac{1}{4(a+b+c)}$$ SOS solution$:$ $$\dfrac{1}{8(a+b+c)}\sum{\dfrac { \left( 52\,{a}^{2}+95\,ab-142\,ac+52\,{b}^{2}-142\...
3
votes
5answers
115 views

Prove that $\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$

If $a+b+c=1$ and $a,b,c>0$ then prove that $$\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$$ My try : We have to prove: $$ab+bc+ca+36abc(ab+bc+ca)\ge 21abc$$ or after homogenising we get : $$\sum a^4b+3\...
2
votes
3answers
116 views

prove that $\sum_{cyc}\frac{{a^2}{b}}{c}\ge a^2+b^2+c^2$ [duplicate]

prove $$\sum_{cyc}\frac{{a^2}{b}}{c}\ge a^2+b^2+c^2$$ where $a,b,c>0$ and $a\ge b\ge c$ My try it seemed qite simple but i couldnt apply the rearrangement inequality directly. so i tried ...
2
votes
5answers
71 views

prove $\sum_{cyc}\frac{a^3}{b}\ge ab+bc+ca$ if $a,b,c>0$

prove $\sum_\text{cyc}\frac{a^3}{b}\ge ab+bc+ca$ if $a,b,c>0$ I couldn't proceed much. I tried rearranging the inequality and it became $a^4c+b^4a+c^4b\ge a^2b^2c+b^2c^2a+c^2a^2b.$ I tried using ...
4
votes
4answers
57 views

prove that $xy+yz+zx\ge x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}$

prove that $xy+yz+zx\ge x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}$ if $x,y,z>0$ My try : dividing inequality by $\sqrt{xyz}$ and putting $\sqrt{x}=a,\sqrt{y}=b,\sqrt{z}=c$ we have to prove $$\sum_{cyc}\frac{...
1
vote
2answers
127 views

Find the smallest possible value of an equation, where $a+b+c=3$

We have the positive real numbers $a, b, c$ such that $a+b+c=3$. Find the minimum value of the equation: $$A=\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}$$ I solved it in the following fashion: $$ \...
2
votes
1answer
93 views

Proving $\sum {\frac {ab}{ \left( a+b \right) ^{2}}}+{\frac {\prod \left( a+b \right) }{16abc}}\geq \frac{5}{4}$

For $a,b,c>0.$ Prove$:$ $${\frac {ab}{ \left( a+b \right) ^{2}}}+{\frac {bc}{ \left( b+c \right) ^{2}}}+{\frac {ac}{ \left( c+a \right) ^{2}}}+\,{\frac { \left( a+b \right) \left( b+c \right) \...
0
votes
1answer
167 views

Proving $3(a^4 + b^4 + c^4 + d^4) + 4abcd \geq (a+b+c+d)(a^3 + b^3 + c^3 + d^3)$

Let $a,b,c,d>0$. Prove that$:$ $$3(a^4 + b^4 + c^4 + d^4) + 4abcd \geqslant (\,a+b+c+d\,)(\,a^3 + b^3 + c^3 + d^{3}\,)$$ As pointed out by @tthnew, this question was posted in: https://...
1
vote
0answers
83 views

Proving $ \sum \frac{a(b+c)}{a^2+bc}+\frac{2\sum (a^2-bc)}{(a+b+c)^2}+\frac{96(a-b)^2(b-c)^2(c-a)^2}{(a+b+c)^6} \leqslant \frac{(a+b+c)^2}{ab+bc+ca} $

For $a,b,c>0.$ Prove$:$ $$\displaystyle \frac{a(b+c)}{a^2+bc}+\frac{b(c+a)}{b^2+ca}+\frac{c(a+b)}{c^2+ab} +\frac{2(a^2+b^2+c^2-ab-bc-ca)}{(a+b+c)^2}+\frac{96(a-b)^2(b-c)^2(c-a)^2}{(a+b+c)^6} \...
1
vote
4answers
123 views

On proving $a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$.

This problem was a "warm-up" problem by the author. Note: $a, b, c$ are non-negative numbers. $$a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$$ I tried to remove the $2$ from ${b+c\...
1
vote
2answers
85 views

Calculate list of all pairs $(x,y)$

The Number Theory question I am trying to solve is :- Write $650$ as a product of irreducible elements in $\mathbb Z[i]$, then list all of the pairs $(x,y) \in$ $Z^2$ with $0 \le x \le y$ such that $x^...
4
votes
1answer
88 views

Prove the inequality $\sum_{cyc}\frac{a^3}{b\sqrt{a^3+8}}\ge 1$

Let $a,b,c>0$ and such $a+b+c=3$,show that $$\sum_{cyc}\dfrac{a^3}{b\sqrt{a^3+8}}\ge 1\tag{1}$$ I tried using Holder's inequality to solve it: $$\sum_{cyc}\dfrac{a^3}{b\sqrt{a^3+8}}\sum b\sum \...

1
2 3 4 5 6