Prove $\big|\langle x,y \rangle\big| \space ≤ \space \lambda \cdot \|x\|^2+\frac{1}{4\lambda} \cdot \|y\|^2$ in an inner product space

Question

I want to prove that if I have an inner product space with $\lambda>0,$ then $$\big|\langle x,y \rangle\big| \space ≤ \space \lambda \cdot \|x\|^2+\frac{1}{4\lambda} \cdot \|y\|^2$$

Where should I begin?

Answer 1

$$(\sqrt{\lambda}||x||-\frac{1}{2\sqrt{\lambda}}||y||)^2 \ge 0 \iff \lambda||x||^2+\frac{1}{4\lambda} ||y||^2 \ge ||x|| \cdot ||y||$$ but $$ ||x|| \cdot ||y|| \ge |\langle x, y \rangle|$$

Answer 2

Note that by AM-GM inequality we have

$$ \lambda \|x\|^2+(1/4\lambda)\|y\|^2\geq \|x\|\|y\|$$

The desired result now follows from the Cauchy-Schwarz inequality.