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Let $\displaystyle f_n=\frac{x^2}{x^2+(1-nx)^2}$ where ($0\le x\le 1,\; n=1,2,3,...$)

Then $|f_n(x)| \le M$.
Find this $M$.

The answer is $1$.
Without any restriction of n, how can we find that bound?

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$0 \leq f_n \leq \frac{x^2+(1-nx)^2}{x^2+(1-nx)^2}$ – N. S. Dec 13 '12 at 4:09
up vote 0 down vote accepted

Let $\frac{x^2}{x^2+(1-nx)^2}=y$

or $x^2(y+yn^2-1)-2nyx+y=0$

As $x$ is real, $(2ny)^2\ge 4y(y+yn^2-1)$

$\implies y^2-y\le 0\implies y(y-1)\le 0\implies 0\le y\le 1$

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May I request to disclose the mistake here? – lab bhattacharjee Feb 7 '13 at 3:10

For all positive integers $n$ and all $x\in[0,1]$, $0\leq x^2\leq x^2+(1-nx)^2>0$. Hence $0\leq\frac{x^2}{x^2+(1-nx)^2}\leq 1$.

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