Want Cauchy and AM-GM-HM

[CmKaN]

I want a Cauchy ,AM-GM problem

P.S. I'm not in Thailand so can't type Thai sorry.

[Lekkoksung]

Where are you now?

[dektep]

1.Let $x,y,z \in R^+$ and satisfy the condition $x+y+z = xyz$ prove that
$xy+yz+zx \geq 3+\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}$

[dektep]

2.Let $x,y,z > 0$ and $xyz=1$ prove that $\sum\frac{1}{x^2+2z^2+3}\leq\frac{1}{2}$

[nooonuii]

3. Find all nonnegative real numbers satisfying the following system of inequalities

$a+b+c\geq 1$

$ab+bc+ca\leq \sqrt{3abc}$

[CmKaN]

maybe some Hint

[dektep]

hint

1.$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \geq 1$
2.$x^2+2z^2+3 = (x^2+z^2)+(z^2+1)+2$

[CmKaN]

Arrrr where is the first Hint come from?

I have a little confuse
$\sum\frac{1}{x^2+2z^2+3}\leq\frac{1}{2}$
It has sum too??

[RoSe-JoKer]

More HinT

2.$x^2+2z^2+3 = (x^2+z^2)+(z^2+1)+2\geq$ $2(xz+z+1)$ then easy done with ...

[CmKaN]

I can't figue out Could you please give solution for the first problem thx

[dektep]

1.$\frac {1}{x^2} + \frac {1}{y^2} + \frac {1}{z^2} \geq \frac {1}{xy} + \frac {1}{yz} + \frac {1}{zx} = 1$

$\rightarrow x^2y^2 + y^2z^2 + z^2x^2 \geq x^2y^2z^2$

$\rightarrow (xy + yz + zx)^2 \geq 2xyz(x + y + z) + x^2y^2z^2 = 3(x + y + z)^2$

$\because (xy + yz + zx - 3)^2 = (xy + yz + zx)^2 - 6(xy + yz + zx) + 9 \geq 3(x^2 + y^2 + z^2) + 9$

$\therefore xy + yz + zx \geq 3 + \sqrt {3(x^2 + y^2 + z^2) + 9} \geq 3 + \sqrt {x^2 + 1} + \sqrt {y^2 + 1} + \sqrt {z^2 + 1}$

[CmKaN]

Where is $\frac {1}{x^2} + \frac {1}{y^2} + \frac {1}{z^2} \geq \frac {1}{xy} + \frac {1}{yz} + \frac {1}{zx} = 1$ from could you please show me ( I'm not good) thanks

[RoSe-JoKer]

เออคือรู้จักอสมการรูปแบบนี้ไหมครับ
$a^2+b^2+c^2\geq ab+bc+ca$
- -* แล้วก็แทน $a=\frac{1}{x}$ แล้ว xy+yz+xz/xyz = 1 ไงครับ
อะไรประมาณนี้อะครับ เอาโจทย์มาเพิ่มนะครับ (อาจได้ลองใช้อสมการนอกจาก Cauchy+AMGMHm หน่อยนะครับแต่คิดว่าไม่น่าจะยาก) ...

[dektep]

โจทย์ของคุณ Rose-Joker ครับ
$1. 5(a^2+b^2+c^2) \leq 6(a^3+b^3+c^3)+1 \leftrightarrow 5(a^2+b^2+c^2)(a+b+c) \leq 6(a^3+b^3+c^3)+(a+b+c)^3$
$\leftrightarrow (b+c-a)(c+a-b)(a+b-c) \leq abc$ ซึงเป็นจริงโดย AM-GM

[dektep]

$2. \frac{(x+y+z)^2}{x^2+xy+y^2} = \frac{1}{1-(ab+bc+ca)-c}$ เมื่อ $a=\frac{x}{x+y+z},b=\frac{y}{x+y+z},c=\frac{z}{x+y+z}$
จะต้องพิสูจน์ว่า $$\frac{1}{1-d-c}+\frac{1}{1-d-b}+\frac{1}{1-d-a} \geq 9$$ เมื่อ $d=ab+bc+ca$ และ $a+b+c=1$
$$\leftrightarrow d(3d-1)^2+(1-4d+9abc) \geq 0$$ ซึ่งคือ Schur's Inequality