Proof of Weierstrass Theorem from Rudin's-Multiple questions though the proof I am studying Stone-Weierstrass theorem. I am not understanding well and I have a multiple point that I need some help to have better understanding.    


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*In the proof, he set $Q_n(x)=C_n (1-x^2)^2$ then $\int _{-1} ^{1} (1-x^2)^n dx \ge 2\int _{0} ^{1/{\sqrt{n}}} (1-x^2)^n \ge 2\int _{0} ^{1/{\sqrt{n}}} (1-n{x^2})dx$
I don't understand why $(1-x^2)^n \ge (1-n{x^2})$.
I understand that the inequality holds but I don't understand how this idea can come up.  

*And there is a part that change of variable is applied according to his explanation but I don't really understand how he got it.
We have $P_n(x)=\int_{-1}^{1}f(x+t)Q_{n}(t)dt$, where $(0 \le x \le 1)$.
Then $P_n(x)=\int_{-x}^{1-x}f(x+t)Q_{n}(t)dt=\int_{0}^{1}f(t)Q_{n}(t-x)dt$. 
I don't understand the process and I don't understand how we can conclude $P_n(x)$ is a polynomial.      
I think understanding in these two parts would help me understand the whole proof.
 A: *

*You can prove $(1-x^2)^n \ge 1-nx^2$ by induction.

*Let $k=x+t$. (I am just calling it $k$, use any other letter besides $x$ or $t$.) Then $t=k-x$. Also, at $t=-x$, we have $k=x+(-x)=0$. At $t=1-x$, we have $k=x+(1-x)=1$. Thus, $$\int_{-x}^{1-x} f(x+t) Q_n(t) \, dt = \int_0^1 f(k) Q_n (k-x) \, dk=\int_0^1 f(t) Q_n (t-x) \, dt.$$
For the right-most equality, replace $k$ with $t$. The value of the definite integral will not change.
A: For 1., just look at $x >0  \mapsto (1-x)^n$ and its second derivative. You will see that it's convex and hence "above its tangent" at $x=0$.
For 2., start by thinking of the case $Q_n(u) = u^k$. If you expand $Q_n(x-t)$, you have a polynomial expression in $x$ and $t$. After multiplication by $f(t)$ and integration on $t$, you're left with a polynomial expression in $x$. 
Another angle on 2., if you know the rules of derivation under the integral sign, you instantly see that if $Q$ is a polynomial of degree $\le n$, the $(n+1)$-th derivative of your integral $P_n(x)$ is nil, which proves it is also a polynomial.
