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

[Albert1]

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$

 

[I like Serena]

Re: prove another inequality

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$

Spoiler

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.$\qquad \blacksquare$

 

[Albert1]

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