Questions tagged [sum-of-squares-method]
Proofs of inequalities by the Sum of Squares method (SOS).
192
questions
2
votes
2answers
48 views
Let $a, b, c>0$. Prove that $\sum \limits_{cyc}{\frac{a}{b+c}\left(\frac{b}{c+a}+\frac{c}{a+b}\right)}\le \frac{(a+b+c)^2}{2(ab+bc+ca)}$
Reducing this whole expression i finally came to this
$$\sum \limits_{cyc}\left(ab^4+a^4b+a^2b^2c\right)\geq \sum \limits_{cyc}\left(a^3b^2+a^2b^3+a^3bc\right)$$
Here I am stuck. I can't prove this.
...
1
vote
1answer
44 views
Prove $P= 7\,{c}^{4}-2\,ab{c}^{2}-2\,ab \left( a+b \right) c+ \left( a+b \right) ^{2} \left( {a}^{2}+{b}^{2} \right) \geqq 0$
For $a,b,c$ are reals$.$ Prove$:$ $$P= 7\,{c}^{4}-2\,ab{c}^{2}-2\,ab \left( a+b \right) c+ \left( a+b \right) ^{2} \left( {a}^{2}+{b}^{2} \right) \geqq 0$$
I found this from Michael Rozenberg's ...
1
vote
0answers
49 views
Prove that $\sum_{\mathrm{cyc}} (40a^6 + 53a^5b) \ge 0$
(P1) Let $a, b, c$ be real numbers. Prove that $40(a^6+b^6+c^6) + 53(a^5b+b^5c+c^5a) \ge 0.$
This inequality is verified by Mathematica. I am particularly interested in (simple) SOS solutions (also ...
3
votes
2answers
46 views
Prove$:$ $\sum\limits_{cyc} (\frac{a}{b+c}-\frac{1}{2}) \geqq (\sum\limits_{cyc} ab)\Big[\sum\limits_{cyc} \frac{1}{(a+b)^2}\Big]-\frac{9}{4}$
For $a,b,c$ are reals and $a+b+c>0, ab+bc+ca>0, (a+b)(b+c)(c+a)>0.$ Prove$:$
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} -\frac{3}{2} \geqq (\sum\limits_{cyc} ab)\Big[\sum\limits_{cyc} \frac{...
5
votes
3answers
123 views
Prove $(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$
For $a,b,c>0; abc=1.$ Prove$:$ $$(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$$
My proof by SOS is ugly and hard if without computer$:$
$$\left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{3}-9\,abc \left( {a}^{3}+...
0
votes
0answers
24 views
Any good usable software or software package for decomposition with constraint?
Does anyone know any good and usable software package, preferably in Windows, that can effectively find SOS (sum of squares) polynomials $s_{0}\left(x\right)$
and $s_{1}\left(x\right)$ for any real ...
1
vote
1answer
40 views
Prove $2\left(x^2+y^2+z^2+1)(x^3y+y^3z+z^3x+xyz\right) \le \left(x^2+y^2+z^2+3xyz\right)^2.$
For $x,y,z\geqq 0$ and $x+y+z=1.$ Prove that$:$
$$2\left(x^2+y^2+z^2+1)(x^3y+y^3z+z^3x+xyz\right) \leqq \left(x^2+y^2+z^2+3xyz\right)^2.$$
Let $y=\hbox{mid} \{x,y,z\}$ and
$\text{P}= \left[\left(x^2+...
1
vote
2answers
59 views
Proving inequality by SOS.
For $x,y,z>0.$ Prove$:$ $$P={x}^{4}y+{x}^{4}z+3\,{x}^{3}{y}^{2}-11\,{x}^{3}yz+3\,{x}^{3}{z}^{2}+3
\,{x}^{2}{y}^{3}+3\,{x}^{2}{y}^{2}z+3\,{x}^{2}y{z}^{2}+3\,{x}^{2}{z}^{
3}+x{y}^{4}-11\,x{y}^{3}z+3\,...
2
votes
4answers
90 views
Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$
Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$
So, using AM-GM, or just pop out squares under square roots we can show:
$$\sqrt{a^2 + ab + b^2} ...
1
vote
1answer
49 views
Find the stronger inequality of $\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}\geq \frac{1}{\sum ab}$
For $a,b,c>0$ and $a+b+c=1.$ Prove$:$ $$\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}\geqq \frac{1}{ab+bc+ca}$$
This inequality is easy and there are two nice proof by AM-GM ...
7
votes
2answers
90 views
Prove $\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geqq 1$
For $x,y,z>0$ and $\sqrt{x} +\sqrt{y} +\sqrt{z} =1.$ Prove that$:$
$$\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geq 1$$
My solution$:$
Let $x=...
2
votes
3answers
72 views
Prove $(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) \leqq 27a^2 b^ 2 c^2$
For $a,b,c>0$$,$ prove$:$ $$(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) \leqq 27a^2 b^ 2 c^2$$
My proof by S-S method$,$ see here.
Another proof by $pqr$ method$:$
Let $p=a+b+c,\,q=ab+bc+ca,\, r=abc.$ This ...
0
votes
1answer
31 views
How to prove Fekete / Markov-Lukasz theorem: nonnegative univariate polynomial on [-1,1] can be decomposed accoording to even/odd-ness of degree
I've been a bit stuck on these few problems. I'm trying to prove the following statements but I'm just not sure what to or how to start:
For a univariate polynomial $f\in \mathbb{R}[x]$ of degree $...
3
votes
4answers
93 views
Prove $\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} +\frac{81abc}{4(a+b+c)^2} \geqq \frac{7}{4} (a+b+c)$
For $a,b,c>0$. Prove that$:$
$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} +\frac{81abc}{4(a+b+c)^2} \geqq \frac{7}{4} (a+b+c)$$
My proof:
We have$:$ $$\text{LHS}-\text{RHS} =\frac{g(a,b,c)}{4abc(a+b+...
3
votes
3answers
61 views
Proving $(a+b+c)^2\prod_{cyc}(a+b)-4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab\geqq 0$
From Mr. Michael Rozenberg solution:
For $a,b,c>0$$,$ prove that$:$
$$(a+b+c)^2\prod_{cyc}(a+b)\geq4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab,$$
I found two SOS proof:
1) $$\text{LHS-RHS}={\frac { \left( ...
0
votes
1answer
84 views
Prove $\sum \sqrt{{\frac {2{a}^{2}b}{a+c}}} \leqq a+b+c$ for $a,b,c>0$
For $a,b,c>0$. Prove that $$\sum \sqrt{{\frac {2{a}^{2}b}{a+c}}} \leqq a+b+c \,\,-----(1)$$
My solution$:$
By C-S, we need to prove: $$(\sum ab) \cdot (\sum \frac{2a}{a+c}) \leqq (a+b+c)^2\, (\...
1
vote
3answers
137 views
Prove $\frac{a(b+c)}{b^2+bc+c^2} \geqslant \frac{8a^2+2a(b+c)-(b-c)^2}{3(a^2+b^2+c^2+ab+bc+ca)}.$
Let $a,\,b,\,c$ are non-negative such that $ab+bc+ca>0.$ Prove that
$$\frac{a(b+c)}{b^2+bc+c^2} \geqslant \frac{8a^2+2a(b+c)-(b-c)^2}{3(a^2+b^2+c^2+ab+bc+ca)}. \quad (1)$$
Note. Because
$$\sum \...
1
vote
1answer
46 views
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 $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le 3+2\...
0
votes
2answers
98 views
Prove $a^2+b^2+c^2 \geqslant \frac{9abc}{a+b+c}+2(1+\sqrt 2)(a-b)(b-c)$
Let $a,b$ and $c$ are positive real numbers. Prove that
$$a^2+b^2+c^2 \geqslant \frac{9abc}{a+b+c}+2(1+\sqrt 2)(a-b)(b-c).$$
My proof is not nice.
Indeed, we need to prove the inequality where $(aā...
2
votes
2answers
70 views
Prove $\frac{a^2}{(b+c)^2}+\frac{b^2}{(c+a)^2}+\frac{c^2}{(a+b)^2}+\frac{1}{4}\ge \frac{a^2+b^2+c^2}{ab+bc+ca}$
For $a,b,c>0$. Prove that: $$\frac{a^2}{(b+c)^2}+\frac{b^2}{(c+a)^2}+\frac{c^2}{(a+b)^2}+\frac{1}{4}\geqq \frac{a^2+b^2+c^2}{ab+bc+ca}$$
NguyenHuyen gave the following expression$:$
$$\sum \...
0
votes
1answer
74 views
Prove $\frac{3}{2} +\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \leqq \frac{a}{b}+\frac{b}{c} +\frac{c}{a}$
For $a,\,b,\,c>0$. Prove: $$\frac{3}{2} +\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \leqq \frac{a}{b}+\frac{b}{c} +\frac{c}{a}$$
My work:
After a lot of caculates, I found:
$\text{RHS-LHS}=$
...
2
votes
3answers
108 views
Proving $(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) +\frac{486(ab+bc+ca)^3}{(a+b+c)^6} \geqq 27$
For $a,b,c > 0$ prove:
$$(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) +\frac{486(ab+bc+ca)^3}{(a+b+c)^6} \geqq 27$$
My work:
I can easy found SOS for it:
$$\text{LHS-RHS}=\sum {\...
4
votes
3answers
114 views
How can I approach this inequality? [duplicate]
Let $a, b$ and $c$ be three non-zero positive numbers. Show that:
$$\sqrt{\frac{2a}{a + b}} + \sqrt{\frac{2b}{b + c}} + \sqrt{\frac{2c}{a + c}} \leq 3$$
I know the triangular inequality would help ...
3
votes
3answers
84 views
show this inequality $\sum_{cyc}\frac{1}{5-2xy}\le 1$
let $x,y,z\ge 0$ and such $x^2+y^2+z^2=3$ show that
$$\sum_{cyc}\dfrac{1}{5-2xy}\le 1$$
try:
$$\sum_{cyc}\dfrac{2xy}{5-2xy}\le 2$$
and
$$\sum_{cyc}\dfrac{2xy}{5-2xy}\le\sum_{cyc}\dfrac{(x+y)^2}{\frac{...
0
votes
1answer
25 views
representation of sum of squares of a globally positive quadratic function
Consider the following quadratic function
\begin{equation}
f(y,z)=1697 y^2+57 y z+81 y+407 z^2-6 z+1
\end{equation}
Using first and second derivatives test, we have shown that the global minimum of $f$...
1
vote
3answers
72 views
Prove $\Big[\sum\limits_{cyc} a(a^2+2bc)\Big]^3 \geqq 3(ab+bc+ca)^2 . \sum\limits_{cyc} a(a^2+2bc)^2$
For $a,b,c>0$, prove that: $$ \Big[\sum\limits_{cyc} a(a^2+2bc)\Big]^3 \geqq 3(ab+bc+ca)^2 . \sum\limits_{cyc} a(a^2+2bc)^2$$
BW works here, but it's very ugly!
My try: Let $p=a+b+c,q=ab+bc+ca,r=...
1
vote
1answer
59 views
Stuck when transforming and solving this
Given abc=1 ( all positive real numbers). Prove that:
$$\frac ab + \frac bc + \frac ca +3( \frac ba +\frac cb +\frac ac) \ge 2(a +b +c+\frac 1a+ \frac 1b +\frac1c)$$
My attempt:
$$\frac ab + \frac ...
1
vote
1answer
60 views
For $a,b,c\ge0$, show $a^2 + b^2 + c^2 = 3$ implies $(a^3 + b^3 + c^3)^2 \geq 3 + 2(a^4 + b^4 + c^4)$.
$a,b,c$ are nonnegative real numbers. If $a^2 + b^2 + c^2 = 3$,
prove that $$(a^3 + b^3 + c^3)^2 \geq 3 + 2(a^4 + b^4 + c^4).$$
I tried using $(a^3 + b^3 + c^3)^2 >= 3(a^3 + b^3 + c^3)abc$ to make ...
1
vote
5answers
68 views
If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is:
If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is:
$A)$ $\le3$ , $B)$ $\ge3$, $(C)$ $\ge2$, $(D)$ $\le2$
My attempt is as follows:-
$$\dfrac{1}{2}\...
4
votes
3answers
111 views
Given $a, b, c>0$, prove $\frac{a^4}{a+b}+\frac{b^4}{b+c}+\frac{c^4}{c+a}\geq \frac{1}{2}(a^{2}c+b^{2}a+c^{2}b)$
Given $a,b,c>0$, prove that $$\frac{a^4}{a+b}+\frac{b^4}{b+c}+\frac{c^4}{c+a}\geq \frac{1}{2}(a^{2}c+b^{2}a+c^{2}b).$$
My attempt:
I have that $$\frac{a^4}{a+b}+\frac{c^2(a+b)}{4}\geq a^{2}c$$
$$\...
1
vote
1answer
99 views
For $a,b,c\in\left[\frac{1}{\sqrt{6}}, 6\right]$: $\sum_{cyc}\frac{4}{a+3b}\geq \sum_{cyc}\frac{3}{a+2b}$
For $a,b,c\in\left[\frac{1}{\sqrt{6}}, 6\right]$ prove that $$\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}.$$
I can't really find a way to exploit the ...
0
votes
2answers
97 views
Prove that $\sqrt{ab+c}+\sqrt{bc+a}+\sqrt{ac+b} \ge 1+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}$ [duplicate]
Prove that $\sqrt{ab+c}+\sqrt{bc+a}+\sqrt{ac+b} \ge 1+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}$
It is given that $a+b+c=1$ and $a,b,c$ are positive real numbers
Then I reduced the inequality to
$\sqrt{(1-a)(1-...
3
votes
3answers
142 views
Proving $a^2 + b^2 + c^2 \geqslant ab + bc + ca$
Let $a, b, c \in \mathbb{R}$. Show that $a^2 + b^2 + c^2 \geqslant ab + bc + ca$.
My reasoning went as follows and I would like to know if it's correct.
$a^2 + b^2 + c^2 \geqslant ab + bc + ca$
$\...
1
vote
4answers
144 views
$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a} \geq \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}$
Prove the following inequality
$$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a} \geq \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}, \enspace \forall a,b,c \in (0,\infty)$$
I tried by multying both sides by the ...
0
votes
1answer
46 views
Prove the following inequality using am gm inequality
Problem:
If $abc \le a+b+c$ then prove that $a^2+b^2+c^2\ge ā3abc$
I squared the inequality to get $a^2+b^2+c^2+2ab+2bc+2ac\ge a^2b^2c^2$. Now I tried to apply am gm but it didn't work.I have to ...
-1
votes
2answers
73 views
Elementary proof for the inequality
I have conjectured the following inequality:
$x+y+z\leq\sqrt{x^2+y^2+z^2}$ where $x,y,z\in \mathbb{R}$
I have tried to come up with an elementary proof but I failed miserably.
Can Anyone help me, ...
0
votes
0answers
15 views
Number systems, squares and probability
first time posting. This has probably been found before, but I found it myself, exciting for me. Sorry if my math language is bad.
My question is about number systems with fractions/benath one.
I ...
1
vote
3answers
113 views
Given positives $x, y , z$ such that $x + y + z = xyz$. Calculate the minimum value of $\frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}$.
Given positives $x, y , z > 1$ such that $x + y + z = xyz$. Calculate the minimum value of $$\large \frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}$$
We have that $x + y + z = xyz \...
3
votes
3answers
211 views
Help with inequality problem [duplicate]
Given $a$ , $b$ , $c \ge 0$ show that
$$\frac{a^2}{(a+b)(a+c)} + \frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)} \ge \frac{3}{4}.$$
I tried using Titu's lemma on it, resulting in
$$\frac{a^2}{(a+b)...
2
votes
1answer
73 views
Prove that for all acute triangles $\triangle ABC$, $r_a + r_b + r_c \ge m_a + m_b + m_c$. [duplicate]
Let $r_b$ and $m_b$ respectively be the exradius of the excircle opposite $B$ and the median drawn from the midpoint of side $CA$ of acute triangles $\triangle ABC$. Prove that $$\large r_a + r_b + ...
4
votes
3answers
105 views
Inequality $\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+x^2} \geqslant \frac{x+y+z}{2}$
Help to prove this Inequality:
If x,y,z are postive real numbers then:
$\dfrac{x^3}{x^2+y^2}+\dfrac{y^3}{y^2+z^2}+\dfrac{z^3}{z^2+x^2} \geqslant \dfrac{x+y+z}{2}$
I tied to use analytic method ...
1
vote
2answers
71 views
How to prove $\sqrt{(a-1)(b-1)}+\sqrt{(a-1)(c-1)}+\sqrt{(b-1)(c-1)}\geq a+b+c+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}$?
I am trying to prove this inequality:
$$\sqrt{(a-1)(b-1)}+\sqrt{(a-1)(c-1)}+\sqrt{(b-1)(c-1)}\geq a+b+c+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}$$
if there is condition $a+b+c=1$. $a, b, c$ are positive ...
2
votes
2answers
161 views
$x,y,z>0$, prove:$\frac{x}{y+z+\sqrt[4]\frac{y^4+z^4}{2}}+\frac{y}{z+x+\sqrt[4]\frac{z^4+x^4}{2}}+\frac{z}{x+y+\sqrt[4]\frac{x^4+y^4}{2}}\geq1$
If $x,y,z>0$, prove:$$ \frac{x}{y+z+\sqrt[4]\frac{y^4+z^4}{2}}+\frac{y}{z+x+\sqrt[4]\frac{z^4+x^4}{2}}+\frac{z}{x+y+\sqrt[4]\frac{x^4+y^4}{2}}\geq1$$ I tried to use classical ineqalities such as AM-...
0
votes
1answer
36 views
Minimizing the cyclic sum $\sum_{\text{cyc}} \sqrt{\frac{a}{b+c}}$ [duplicate]
I want to minimize the following cyclic sum, where $a,b,c>0$: $$\sum_{\text{cyc}} \sqrt{\frac{a}{b+c}}=\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}.$$
More precisely, I think ...
0
votes
1answer
62 views
Prove that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ $\geq$ $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ [duplicate]
Let $a,b,c$ be the sides of a triangle and $a+b+c=3$
Prove that:
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
My Attempt
I think this inequality could be proved ...
1
vote
2answers
92 views
Prove the given inequality if $a+b+c=1$
Let $a,b,c$ be positive real numbers such that $a+b+c=1$, then prove that
$\frac{a}{a^2 +b^3 +c^3}+\frac{b}{a^3 +b^2 +c^3}+\frac{c}{a^3 +b^3 +c^2} \leq \frac{1}{5abc}$
Please provide some hint to ...
6
votes
2answers
219 views
Inequality : $\frac{a}{\exp(a+b)}+\frac{b}{\exp(b+c)}+\frac{c}{\exp(c+a)}\leq \exp\Big(\frac{-2}{3}\Big)$
It's a charming problem :
Let $a,b,c>0$ such that $a+b+c=1$ then we have :
$$\frac{a}{\exp(a+b)}+\frac{b}{\exp(b+c)}+\frac{c}{\exp(c+a)}\leq \exp\Big(\frac{-2}{3}\Big)$$
I know the identity :
...
1
vote
0answers
62 views
Show that every nonnegative polynomial $P \in \mathbb R [x]$ can be written as a sum of squares of real polynomials
Define a real polynomial $P \in \mathbb R[x]$ to be nonnegative if $P(x) \geq 0$ for all $x \in \mathbb R$.
Show that every nonnegative polynomial $P \in \mathbb R [x]$ can be written as a sum of ...
1
vote
1answer
82 views
Nesbitt by Nesbitt
The title not says I'm Nesbitt but just only says it's a refinement of Nesbitt's inequality by itself so we have :
Let $a,b,c>0$ and $a\geq b \geq c$ then we have :
$$\frac{a}{b+c}+\frac{b}{a+...
0
votes
1answer
92 views
Inequality with square roots conditions
Let $a,b,c>0$ and $a+b+c=9$.
Prove that: $$(\sqrt{a}+\sqrt{b}+\sqrt{c})^2\geq ab+bc+ca.$$
(Edit) I'm sorry. The problem should have beeen:
Let $a,b,c\geq1$ and $a+b+c=9$.
Prove that: $$(\sqrt{a}+\...
