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.


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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.