Can the Inequality $x,y,z>1$ be Proven with a Hint of 48?

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SUMMARY

The inequality $\dfrac{x^4}{(y-1)^2}+\dfrac{y^4}{(z-1)^2}+\dfrac{z^4}{(x-1)^2}\geq 48$ for $x, y, z > 1$ can be proven using techniques from inequality theory and the Cauchy-Schwarz inequality. The discussion emphasizes the necessity of manipulating the terms effectively to establish the lower bound of 48. Participants suggest leveraging the AM-GM inequality as a potential approach to simplify the proof process.

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  • Study the Cauchy-Schwarz inequality in depth to understand its applications in proving inequalities.
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Albert1
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$x,y,z>1$

please prove :

$\dfrac{x^4}{(y-1)^2}+\dfrac{y^4}{(z-1)^2}+\dfrac{z^4}{(x-1)^2}\geq 48$
 
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Albert said:
$x,y,z>1$

please prove :

$\dfrac{x^4}{(y-1)^2}+\dfrac{y^4}{(z-1)^2}+\dfrac{z^4}{(x-1)^2}\geq 48$
hint:
prove :$\dfrac{x^4}{(y-1)^2}\geq 32(x-y)+16$
 
Albert said:
hint:
prove :$\dfrac{x^4}{(y-1)^2}\geq 32(x-y)+16$
$\dfrac{x^4}{(y-1)^2}+16(y-1)+16(y-1)+16\geq 4\sqrt[4]{16^3x^4}= 32x$
or :$\dfrac{x^4}{(y-1)^2}\geq 32(x-y)+16$
please complete the rest
 

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