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?
 A: Hint: We have $$0\le (x^2-5)^2$$ and by expanding $$10x^2\le x^4+25$$ so $$\frac{10x^2}{x^4+25}\le 1$$
A: Since $x^2\ge 0$ and $x^4+25>0$ we have
$$y=\frac{x^2}{x^4+25}\ge 0$$
and since $y$ is bounded above
$$\frac{x^2}{x^4+25}\le k \implies kx^4-x^2+25k \ge 0$$
$$\Delta=1-100k^2=0 \implies k=\frac1{10}$$
A: From the AM-GM Inequality, we have
$$\begin{align}
x^4+25&\ge  2\sqrt{25x^4}\\\\
&=10x^2\tag1
\end{align}$$
Applying $(1)$ reveals  
$$0\le \frac{x^2}{x^4+25}\le \frac{x^2}{10x^2}=\frac1{10}$$
And we are done!
A: $x^2, x^4, x^4 + 25 \ge 0$ 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.
A: $$\frac{x^2}{x^4+25}=\le\frac1{10}\iff x^4-10x^2+25\ge0$$
Can you now factor the rightmost expression and prove the inequality is true?
A: Consider the function
$$f(x)=\frac{x^2}{x^4+25}\implies f'(x)=-\frac{2 x \left(x^4-25\right)}{\left(x^4+25\right)^2}\implies f''(x)=\frac{2 \left(3 x^8-300 x^4+625\right)}{\left(x^4+25\right)^3}$$ The first derivative cancels at $x=0$ and the second derivative test shows that this is a minimum. Thos other roots are $x_\pm=\pm \sqrt 5$ which corresponds to maximum values of the function. Now, compute $f(\pm \sqrt 5)$ to get the result.
