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Q: Prove that for arbitrary $x_1,x_2,...,x_n$:

$$\frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+...+\frac{x_n}{1+x_1^2+x_2^2+...+x_n^2}\lt\sqrt{n}$$

I tried using mathematical induction. The case for $n=1$ is obvious. Assume it is true for $n=k$. So for $n=k+1$:

$$\frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+...+\frac{x_n}{1+x_1^2+x_2^2+...+x_n^2}+\frac{x_{n+1}}{1+x_1^2+x_2^2+...+x_{n+1}^2}\lt\sqrt{n+1}$$

$$\sqrt{n}+\frac{x_{n+1}}{1+x_1^2+x_2^2+...+x_{n+1}^2}\lt\sqrt{n+1}$$ by the induction hypothesis

$$\frac{x_{n+1}}{1+x_1^2+x_2^2+...+x_{n+1}^2}\lt\sqrt{n+1}-\sqrt{n}$$

$$(\sqrt{n+1}+\sqrt{n})x_{n+1}\gt{1+x_1^2+x_2^2+...+x_{n+1}^2}$$

How do I proceed from here? Am I even on the right track?

Vikram
  • 196

1 Answers1

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No, I don't think you are on the right track. You want to prove $$\frac{x_{n+1}}{1+x_1^2+x_2^2+...+x_{n+1}^2}\lt\sqrt{n+1}-\sqrt{n}$$ Ask yourself if this can possibly be true. To make the left-hand side large, we can take $$ x_1=x_2=\dots x_n=0$$ Now we need to prove $$\frac{x}{1+x^2}<\sqrt{n+1}-\sqrt{n}\tag{1}$$ for arbitrary $x$ and all integers $n>1$. This is false. Take any positive value of $x$. Then the left-hand side 0f $(1)$ is some positive constant, but the right-hand side goes to $0$ as $n\to\infty,$ and $(1)$ is false.

In fact, it's enough to take $x=1,\space n=1.$

saulspatz
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