Questions tagged [sum-of-squares-method]

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

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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. ...
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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 ...
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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 ...
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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
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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}+...
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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 ...
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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+...
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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\,...
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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
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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 ...
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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=...
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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 ...
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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 $...
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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+...
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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( ...
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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\, (\...
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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 \...
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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\...
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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 $(a−...
2
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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 \...
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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
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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
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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 ...
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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{...
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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 $f$...
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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=...
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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
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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 ...
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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
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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
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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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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
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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
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1answer
46 views

Prove the following inequality using am gm inequality

Problem: If $abc \le a+b+c$ then prove that $a^2+b^2+c^2\ge √3abc$ 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
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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
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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 ...
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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
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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
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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
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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
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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}+\...