We have `(5/4)^(1-x) gt (16/25)^(2(1+sqrt(x))` or `(4/5)^(x-1) gt (4/5)^(4(1+sqrt(x))`
Since the base `0 gt 4/5 gt 1`, the inequality to the inequality `x-1 gt 4(1+sqrt(x))`
`rArr (x-5)/4 gt sqrt(x)`
Now, RHS is positive
`rArr (x-5)/4 gt rArr x gt 5`.........(i)
We have `(x-5)/4 gt sqrt(x)`
both sides are positive, so squaring both side
`rArr (x-5)^(2)/16 gt x` or `(x-5)^(2)/16 -x gt 0`
or `x^(2)-26x+25 gt 0` or `(x-25)(x-1) gt 0`
`rArr x in (-infty,1) cup (25, infty)`.................(ii)
intersection (i) and (ii) gives `x in (25,infty)`
