<s>[INST] If $y=\frac{x^2}{x^4+25}$ ,prove: $0 \leq y \leq \frac{1}{10}$ We know that:
$$y=\frac{x^2}{x^4+25}$$
Then we have to prove that:
$$0 \leq y \leq \frac{1}{10}$$
How to use what we obtain from first fraction to prove what it wanted? Do you have an easy idea?
 [/INST] $x^2, x^4, x^4 + 25 \ge 0$ so $\frac {x^2}{x^4 + 25} \ge 0$.
If $x = 0$ then $\frac {x^2}{x^4 + 25} = 0 < \frac 1{10}$. 
If $x \ne 0$ then $\frac {x^2}{x^4 + 25} \le \frac 1{10}\iff \frac {x^4 + 25}{x^2} \ge 10$.
And by AM-GM... $\frac {x^4 + 25}{x^2}  = x^2 + \frac {25}{x^2} \ge 2\sqrt{x^2*\frac {25}{x^2}}= 2\sqrt{25} =10$.
....  But perhaps the most direct way as all terms are positive.
$\frac {x^2}{x^4 + 25} \le \frac 1{10} \iff 10x^2  \le x^4 + 25 \iff x^4 -10x^2 + 25=(x^2 -5)^2 \ge 0$.  which must be true as $(x^2 - 5)^2$ is a perfect square.
[Note:  the second argument is essentially replicating a prove of the AM-GM: Namely for positive $a,b$ we know $a + b \ge 2\sqrt{ab}\iff a-2\sqrt{ab} + b = (\sqrt a - \sqrt b) \ge 0$  which.... must be true.
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