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I have a problem that I can not solve although I know that the statement is true.

For all $t\ge0$ the real root of $x(x+t)^2-4=0$ is greater than or equal to the largest real root of $4x^6-6x^4+4t^3x^3+t^6-3t^4=0$. Does anyone have some idea to solve this problem?

Kyle
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Piquito
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  • are you sure that you have made no typo? – Dr. Sonnhard Graubner Mar 22 '18 at 15:26
  • Yes I do. An easy way to see it: you can check this graphically. – Piquito Mar 22 '18 at 15:47
  • "Checking it graphically" pretty much constitutes a proof: Show that $x(x+t)^2-4$ is still negative at a value such that the other polynomial is guaranteed to be positive there and everywhere larger. – Paul Sinclair Mar 23 '18 at 00:20
  • Not really I think. For example the difficult inequality proposed by @Andreas $$\frac{a^3+b^3+c^3}{3}\geq\sqrt{\frac{a^4+b^4+c^4}{3}}$$ for non-negative with the constraint $ (a + b) (a + c) (b + c) = 8 $ can be easily verified graphically but this is not what is being asked. – Piquito Mar 23 '18 at 00:47
  • Equivalence inequaty is $(t^3+2x^3)^2/(t^4+2x^4)>3$ – Takahiro Waki Mar 23 '18 at 02:51

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