App of deriviatives

last one is nice, but it appears to be true for x>1 (not x>0) as you have typed

from the given equation we get x^7-x^5 + x^3-x = 3a \Rightarrow (x^2-1)(x^5+x) = 3a \Rightarrow x^2-1 = \frac{3a}{x^5+x}

Now we have to prove that x⁸>1+6a or equivalently x⁸-1>6a

x^8 -1 = (x^2-1)(x^6+x^4+x^2+1) = \frac{3a}{x^5+x}(x^6+x^4+x^2+1)

Hence it suffices to prove that \frac {x^6+x^4+x^2+1}{x^5+x}> 2 or x^6+x^4+x^2+1 > 2x^5+2x

which is true by AM-GM as x⁶+x⁴>2x⁵ and x²+1>2x (strict ineq because x>1)

or could write x^6-2x^5 + x^4+x^2-2x+1 = (x^3-x^2)^2 + (x-1)^2 >0to prove the same thing.