Is the substitution of variables allowed in proofs?

Question

EDIT: I later realized no one saw that substituting $\sqrt{x} = x$ is not valid because they are not equal, you would have to assume $x$ as a completely different variable (i.e. $a$) and not the same one because they have no relation to each other. However, the solution to my proof problem is given.

So for example I have to prove:

$2y \leq \frac{y^2}{x^2} + x^2$ (for any real numbers $x$ and $y$ and $x$ $\neq$ $0$),

and I do some rough work at start with this and end up with the arithmetic-geometric mean inequality:

$\sqrt{xy} \le \frac{x+y}{2}$ (using substitution of variables).

Now to prove it, I start with the arithmetic-geometric mean inequality, then do the following steps:

$\sqrt{xy} \leq \frac{x+y}{2}$

$\sqrt{x}\sqrt{y} \leq \frac{x+y}{2}$

Now, substitute $\sqrt{x}$ with $x$ and so:

$xy \leq x^2+\frac{y^2}{2}$

Then throughout, I do one more substitution:

$\sqrt{x} = x$

to end up with the first (1) equation. Would this be allowed or is there another way of proving it? Such as squaring the right side of arithmetic-geometric mean inequality? I am not too sure. Thanks.

Answer 1

HINT

We have that

$$2y\le \frac{y^2}{x^2}+x^2 \iff2yx^2\le y^2+x^4 \iff x^4-2yx^2+y^2\ge 0$$

Answer 2

$x^2 > 0$ we have

$2y \le \frac {y^2}{x^2} + x^2 \iff yx^2 \le \frac{x^2 + x^4}2$.

Now do the substition. Substitute $y = \sqrt a$ and $x^2 = \sqrt b$.

EXCEPT we don't know that $y$ is positive.

Let $y = \pm \sqrt a$ and $x^2 = \sqrt b$ so $y^2 = a \ge 0$ and $b = x^4 > 0$

And we know $\sqrt ab \le \frac {a+b}2$ for $a,b \ge 0$ so

$\sqrt {y^2} \sqrt {x^4} \le \frac {y^2 + x^4}2$ so

$|y|x^2 \le \frac {y^2 + x^4}2$ so

$2|y| \le \frac {y^2}{x^2} + x^2$ and finally

$2y \le 2|y| \le \frac {y^2}{x^2} + x^2$.

(remember $k \le |k|$ becuase if $k < 0$ then $k < 0 < |k|$. And if $k \ge 0$ then $k = |k|$.)