2

let $x,y,z>0$,and such $$x+y+z=3$$ prove or disprove this $$(x^2+2)(y^2+2)(z^2+2)\ge (\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}+\sqrt{x^2+xy+y^2})^2\tag{1}$$

I know this well know inequality 1 $$(a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2$$

and we all know this inequality 2 $$(\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}+\sqrt{a^2+ab+b^2})^2\ge 3(a+b+c)^2$$

poof 1: since $$(b-c)^2+2(bc-1)^2\ge 0$$ so $$(b^2+2)(c^2+2)\ge 3\left(1+\dfrac{(b+c)^2}{2}\right)$$

Use Cauchy-Schwarz inequality ,we have $$(a^2+2)\left(1+\dfrac{(b+c)^2}{2}\right)\ge (a+b+c)^2$$ By done!

proof 2: since $$\sqrt{b^2+bc+c^2}=\sqrt{\dfrac{3}{4}(b+c)^2+\dfrac{1}{4}(b-c)^2}\ge\dfrac{\sqrt{3}}{2}(b+c)$$ so $$\sum_{cyc}\sqrt{a^2+ab+b^2}\ge\sqrt{3}(a+b+c)$$ then $$(\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}+\sqrt{a^2+ab+b^2})^2\ge 3(a+b+c)^2$$ By done!

But my inequality (1) is stronger this two inequality.Thank you

math110
  • 93,304

1 Answers1

3

$(\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}+\sqrt{x^2+xy+y^2})^2 \le 3(y^2+yz+z^2+z^2+zx+x^2+z^2+zx+x^2) \iff $

$(x^2+2)(y^2+2)(z^2+2)- 3(y^2+yz+z^2+z^2+zx+x^2+z^2+zx+x^2) \ge0 \iff x^2y^2z^2+2y^2z^2+2x^2z^2-2z^2-3yz-3xz+2x^2y^2-2y^2-3xy-2x^2+8\ge 0 $$ \tag 2$

let $3u=x+y+z,3v^2=xy+yz+xz,w^3=xyz,\to u=1\ge v \ge w$,

$(2)$ becomes:

$ w^6-24v^2w^3+3v^2+20 \ge 0 \iff w^6-24 w^3+3w^2+20\ge 0$ $ (3)$

$w^6+w^2+1\ge 3w^{\frac{8}{3}} \ge 3 w^3, 19+2w^2 \ge 21w^3 \implies $ $(3)$ is true.

the "=" will hold when $v=w=1=u \implies x=y=z=1$

QED.

chenbai
  • 7,581