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I am learning about the Cauchy-Schwarz inequality and I cam across this question:

Consider the function $f(x) = \frac{(x+k)^2}{x^2 +1}$ where $k>0$ and $x$ is a real number. Show that $f(x)\leq k^2 +1$ for all $x$ and $k>0$ using the Cauchy-Schwarz inequality.

I have tried to use the integral definition of the inequality but that got me no where. I am not very versed with using mathematical induction yet either, but I can understand proofs that use it. I proved this at first by using calculus and then showing that the maximum was less than $k^2 + 1$, but I want to know how to use C-S inequality to solve this problem.

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$$(1,x)\cdot(k,1)=x+k,\hbox{ thus}$$ $$(x+k)^2= ((1,x)\cdot(k,1))^2\le (1,x)^2\cdot (1,k)^2=(1+x^2)\cdot (1+k^2)$$ P.S. inequality is in vector form with dot product $|(\mathbf{u}\cdot \mathbf{v})|\le |\mathbf{u}|\cdot|\mathbf{v}|$.

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  • $\begingroup$ Thanks man. This really helped. $\endgroup$
    – C Squared
    Jul 14 '20 at 22:39

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