Question

Prove that inequality:
${|z_1+z_2|}^2\le (1+\lambda ){\left|z_1\right|}^2+(1+\frac{1}{\lambda }){\left|z_2\right|}^2$
where $\lambda >0$.

Answer 1

${|z_1+z_2|}^2=(z_1+z_2)\left(\overline{z_1+z_2}\right)$
$=(z_1+z_2)(\overline{z_1}+\overline{z_2})$
$={\left|z_1\right|}^2+{\left|z_2\right|}^2+z_1\overline{z_2}+\overline{z_1\overline{z_2}}$
$={\left|z_1\right|}^2+{\left|z_2\right|}^2+2R\left(z_1{\overline{z}}_2\right)$
$\le {\left|z_1\right|}^2+{\left|z_2\right|}^2+2\left|z_1\overline{z_2}\right|$
$={\left|z_1\right|}^2+{\left|z_2\right|}^2+2\left|z_1\right|\left|z_2\right|$ ......$\left(1\right)$
Now $\lambda >0\therefore \sqrt{\lambda }$ is real.
We write the last term in $\left(1\right)$ as
$2\left|\sqrt{\lambda }{z}_1\right|\left|\frac{1}{\sqrt{\lambda }}{z}_2\right|$.$=2AB\le A^2+B^2$
$=\lambda {\left|z_1\right|}^2+\frac{1}{\lambda }{\left|z_2\right|}^2$. Put in $\left(1\right)$ etc.