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# JOMO 6, Long 2

Given that for $$x, y, z\geq 0$$, we have $$xy+yz+zx = 3$$ Then prove that: $\frac{x+y+z}{x^2y^2z^2} \geq \frac{9}{x^3+y^3+z^3 - (x+y+z)\left[(x+1)(x-1)+(y+1)(y-1)+(z-1)(z+1)\right]}$

Note by Yan Yau Cheng
3 years, 3 months ago

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Note that: x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx) Then, x^3+y^3+z^3-3xyz=(x+y+z)[(x^2-1)+(y^2-1)+(z^2-1)] So, x^3+y^3+z^3-(x+y+z)[(x+1)(x-1)+(y+1)(y-1)+(z+1)(z-1)]=3xyz Now, it simply to prove (x+y+z)/[(xyz)^2] >= 9/(3xyz)=3/(xyz) <=> (x+y+z)/(xyz) >= 3 <=> 1/(xy)+1/(yz)+1/(xz) >= 3 Since x,y,z>0, by AM-HM, (xy+yz+zx)[1/(xy)+1/(yz)+1/(xz)] >= 9 => [1/(xy)+1/(yz)+1/(xz)] >= 3 , and we done~ :) ~

- 3 years, 3 months ago

$${\LaTeX}$$'d

Note that: $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$

Then, $$x^3+y^3+z^3-3xyz=(x+y+z)[(x^2-1)+(y^2-1)+(z^2-1)]$$

So, $$x^3+y^3+z^3-(x+y+z)[(x+1)(x-1)+(y+1)(y-1)+(z+1)(z-1)]=3xyz$$

Now, it simply to prove \begin{align*}\dfrac{x+y+z}{(xyz)^2} &\ge \dfrac{9}{3xyz}=\dfrac{3}{xyz}\\ &\iff \dfrac{x+y+z}{xyz} \ge 3\\ &\iff \dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz} \ge 3 \end{align*} Since $$x,y,z > 0$$, by AM-HM, \begin{align*}(xy+yz+zx)\left[\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz} \right] &\ge 9 \\ \implies \dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz} &\ge 3\end{align*} and we done~ :) ~

- 3 years, 2 months ago