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$x\ge -\dfrac{1}{2}; \dfrac{x}{y}>1$

Find the minimum of the function $\frac{2x^{3}+1}{4y\left ( x-y \right )}$

I thought of using a few values of $x$ and $y$, but that seemed extremely inefficient. I've seen many people describing AM-GM and Cauchy-Schwarz, but I'm not sure if that even applies to this problem. I have reinforced my understanding of AM-GM and Cauchy-Schwarz, but I'm still not sure how to solve this. Can someone guide me through this?

A Piercing Arrow
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  • You might want to check your variables, $x,y,a$ and $b$? – Teddy38 Dec 03 '17 at 17:16
  • What do $a$ and $b$ have to do with $x$ and $y$. Are there any restrictions put on $a$ and $b$? – kimchi lover Dec 03 '17 at 17:16
  • You can equate the partial derivatives to zero and get a critical point (1,0.5). To see if it is a max or min or neither let $ X = 1 + h $ , $ Y = 1/2 + k $ where $ h, k $ are small. Expand using Binomial Theorem to second order in $ h, k $. Bring terms in $ h^{2}, k^{2} $ and $ hk $ together above the line. These will give a positive definite quadratic form, showing that the expression can only increase from a minimum of 3 as $ h $ and $ k $ increase or decrease. Note $ (1 + h)^{-1} $ ~ $ 1 - h + h^{2}..$. – kieran Dec 04 '17 at 06:03

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prove that $$\frac{2a^3+1}{4b(a-b)}\geq 3$$ and the equal sign holds if $$a=-\frac{1}{2},b=-\frac{1}{4}$$

Dr. Sonnhard Graubner
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