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If $x, y, z$ are positive real numbers such that $$xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}},$$ then prove that $x+y+z\geq \sqrt{3}$.

alexwlchan
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novak
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3 Answers3

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Taking the contrapositive, we suppose $x+y+z<\sqrt 3$ and want to prove $$xy+yz+zx<\frac 1{\sqrt{x^2+y^2+z^2}}\iff (xy+yz+zx)^2(x^2+y^2+z^2)<1$$ Scale $(x,y,z)$ by dividing all variables by $\sqrt3$. Then $x+y+z<3$ and we want to prove $$(xy+yz+zx)^2(x^2+y^2+z^2)<27$$

And that's a simple AM-GM: $$(xy+yz+zx)^2(x^2+y^2+z^2)\le\left(\frac{2(xy+yz+zx)+x^2+y^2+z^2}3\right)^3=\left(\frac{(x+y+z)^2}3\right)^3<27$$

user2345215
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  • i could not understand after the scale(x,y,z) part. Please help @user2345215 – novak Jan 21 '15 at 12:27
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    @novak Actually, it isn't really necessary. You can stay with $x+y+z<\sqrt 3$ and then compute $((x+y+z)^2/3)^3<((\sqrt3)^2/3)^3=1$ at the end. I used it as an intermediate step when finding a solution to get rid of the square root of 3. – user2345215 Jan 21 '15 at 12:39
  • Thanks just realized its easy AM-GM – novak Jan 30 '15 at 09:50
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$(x+y+z)^2 \geq 3 \iff x^2+y^2+z^2 + 2(xy+yz+zx) \geq 3 \iff x^2+y^2+z^2+ \dfrac{2}{\sqrt{x^2+y^2+z^2}}\geq 3 \iff t^2 + \dfrac{2}{t} \geq 3$. By AM-GM inequality: $t^2+\dfrac{2}{t} =t^2+ \dfrac{1}{t}+\dfrac{1}{t}\geq 3$.

DeepSea
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  • I'm sorry if my solution is the same as yours, but I assure you I couldn't understand it at all and noticed some similarity only after finishing my proof. – user2345215 Jan 20 '15 at 22:01
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enter image description here

Sketch indicates the truth of the proposition. Blue boxes : $x+y+z > \sqrt{3}$. Blue with purple centers $$xy + yz + zx > \frac{1}{\sqrt{x^2+y^2+z^2}}$$ The surfaces indicate where equality holds.

Alan
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