MHB Prove Inequality: $x^4,y^4,z^4 \geq 48(y-1)^2(z-1)^2(x-1)^2$

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The discussion focuses on proving the inequality \( \frac{x^4}{(y-1)^2} + \frac{y^4}{(z-1)^2} + \frac{z^4}{(x-1)^2} \geq 48 \) under the conditions \( x > 1, y > 1, z > 1 \). The proof involves manipulating the terms to show that each fraction is greater than or equal to a certain expression involving the differences between the variables. Specifically, inequalities are established for each term, leading to a combination that satisfies the original inequality. The final conclusion is reached by summing these inequalities. This demonstrates the validity of the proposed inequality under the given conditions.
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x>1,y>1 and z>1

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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Re: prove another inequality

Albert said:
x>1,y>1 and z>1

prove :$\dfrac {x^4}{(y-1)^2}+\dfrac {y^4}{(z-1)^2}+\dfrac

{z^4}{(x-1)^2}\geq 48$

If any of the coordinates is close to either $1$ or $\infty$, the LHS approaches $\infty$.
Therefore there must be a minimum where all coordinates are between $1$ and $\infty$.

Due to the cyclic symmetry, the optimum must have equal coordinates, meaning $x=y=z$.
Substituting gives us an LHS of
$$\frac {3x^4}{(x-1)^3}$$
Taking the derivative, setting it to zero, and solving, yields $x=y=z=2$.
The corresponding minimum is 48.$\qquad \blacksquare$
 
Re: prove another inequality

$\,\,\dfrac {x^4}{(y-1)^2}+16(y-1)+16(y-1)+16\geq 4\sqrt[4]{16^3x^4}=32x$

$\therefore \,\,\dfrac {x^4}{(y-1)^2}\,\, \geq 32(x-y)+16---(1)$$\therefore \,\,\dfrac {y^4}{(z-1)^2}\,\, \geq 32(y-z)+16---(2)$

$\therefore \,\,\dfrac {z^4}{(x-1)^2}\,\, \geq 32(z-x)+16---(3)$

(1)+(2)+(3) and the proof is accomplished
 
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