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

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The discussion revolves around proving the inequality involving positive variables x, y, and z, each greater than 1. The specific inequality to be proven is that the sum of the fractions, $\dfrac{x^4}{(y-1)^2}+\dfrac{y^4}{(z-1)^2}+\dfrac{z^4}{(x-1)^2}$, is greater than or equal to 48. Participants are encouraged to explore the hint provided, which suggests a potential method or insight for the proof. The focus remains on finding a rigorous mathematical approach to establish the inequality under the given conditions. The conversation emphasizes the need for clarity in the proof process and the importance of the hint in guiding the solution.
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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
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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