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let $x,y,z>0$,and such $$x^2y^2+y^2z^2+x^2z^2+2x^2y^2z^2=1$$ show that $$x+y+z\ge\dfrac{3\sqrt{2}}{2}$$

My idea: in $\Delta ABC$,we have $$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$ so let $$xy=\cos{A},yz=\cos{B},zx=\cos{C},A,B,C\in(0,\dfrac{\pi}{2})$$ $$\Longleftrightarrow \sqrt{\dfrac{\cos{B}\cos{C}}{\cos{A}}}+\sqrt{\dfrac{\cos{A}\cos{C}}{\cos{B}}}+\sqrt{\dfrac{\cos{A}\cos{B}}{\cos{C}}}\ge\dfrac{3\sqrt{2}}{2}$$ But I can't.Thank you very much

Michael Rozenberg
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  • The problem only makes sense if there is a restriction that $\Delta ABC$ is acute. Is this stated in the original problem? – Marconius Jul 09 '15 at 18:55

1 Answers1

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The claim is false: If we let $x = 0$, $y=1$ and $z=1$, then we have $x+y+z = 2 < 3 \sqrt{2} / 2$. (This conclusion holds by continuity for some $x,y,z > 0$, too.)

J. J.
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  • But then if $x \rightarrow 0^+$ this means wlog that $cos B \rightarrow 0^+$, so then the second term (and hence the sum) $\gg 3\sqrt2 / 2$. If on the other hand, $x,y,z$ are the operands of the square roots, then two must simultaneously tend to zero. – Marconius Jul 09 '15 at 18:37