Inequality for amateurs I have a new inequality with two variables this is the following :

Let $82\geq y\geq 2$ and $0\leq x \leq 1$ then we have :
  $$\frac{(1+x)^{0.25}}{(y-1)^{0.25}+(1-x)^{0.25}}+\frac{(1-x)^{0.25}}{(y-1)^{0.25}+(1+x)^{0.25}}+\frac{(y-1)^{0.25}}{2}\geq 1.5$$

I have tried differents methods but it fails always . Furthermore I can't prove that if we fix $y$ the funtion with $x$ is increasing for $x\in [0;1]$
but I have this :
Remark that the inequality is equivalent to :
$$f(x)+f(-x)+\frac{(y-1)^{0.25}}{2}\geq 1.5$$
With :
$$f(x)=\frac{(1+x)^{0.25}}{(y-1)^{0.25}+(1-x)^{0.25}}$$
It remains to prove that :
$$f(x)\geq f(-x)$$
Wich is true so the function $g(x)=f(x)+f(-x)$ is increasing but I'm not sure that my arguments are correct .
Can someone help me ? 
 A: $$\frac{(1+x)^{0.25}}{(y-1)^{0.25}+(1-x)^{0.25}}+\frac{(1-x)^{0.25}}{(y-1)^{0.25}+(1+x)^{0.25}}+\frac{(y-1)^{0.25}}{2}\geq 1.5\space\space.....(*)$$
I have solved this problem with an arduous and long solution but suddenly I realized that I can also reason as follows:
Put $$h_a(x)=\frac{(1+x)^{0.25}}{a+(1-x)^{0.25}}\\k_a(x)=\frac{(1-x)^{0.25}}{a+(1+x)^{0.25}}$$ where $a$ is a parameter equal to $(y-1)^{0.25}$ with $1\le a\le 3$.
$h_a(x)$ is increasing from $h_a(0)=\dfrac{1}{1+a}\space\text {to }\space 
 h_a(1)=\dfrac{2^{0.25}}{a}$ and $k_a(x)$ is decreasing from $k_a(0)=\dfrac{1}{1+a}\space\text {to }\space k_a(1)= 0$ so both functions are positive.  ($h_a(x)$ and $k_a(x)$ are above and below the horizontal line $y=\dfrac{1}{1+a}$ respectively but they are not symmetrical with respect to this line).
Since both functions are positive and $h_a(0)=k_a(0)=\dfrac{1}{1+a}$, the minimum of $h_a(x)+k_a(x)$ is clearly equal to $\dfrac{2}{1+a}$.
Now we need to prove $$h_a(x)+k_a(x) +\frac a2\ge 1.5$$ $$h_a(x)+k_a(x) +\frac a2\ge \dfrac{2}{1+a}+\frac a2\ge 1.5\iff (a-1)^2\ge 0\iff a\ge 1$$ 
Our values for $a$ are such that $1\le a\le 3$ and $a=1$ is the only in this interval $[1,3]$ for which the above inequality $(*)$ becomes equality.
