Question

If $Z_1$ and $Z_2$ are two complex numbers and $c>0,$ then prove that
${|z_1+z_2|}^2\le (1+c){\left|z_1\right|}^2+(1+c^{-1}){\left|z_2\right|}^2$

Answer 1

We have to prove:

${|z_1+z_2|}^2\le (1+c){\left|z_1\right|}^2+(1+c^{-1}){\left|z_2\right|}^2$

i.e. ${\left|z_1\right|}^2+{\left|z_2\right|}^2+z_1{\overline{z}}_2+{\overline{z}}_1{z}_2\le (1+c){\left|z_1\right|}^2+(1+c^{-1}){\left|z_2\right|}^3$

or $z_1{\overline{z}}_2+{\overline{z}}_1{z}_2\le c{\left|z_1\right|}^2+c^{-1}{\left|z_2\right|}^2$

or $c{\left|z_1\right|}^2+\frac{1}{c}{\left|z_2\right|}^2-z_1{\overline{z}}_2-{\overline{z}}_1{z}_2\ge 0$ (using Re$\left(z_1{\overline{z}}_2\right)\le \left|z_1{\overline{z}}_2\right|$)

or ${(\sqrt{c}\left|z_1\right|-\frac{1}{\sqrt{c}}\left|z_2\right|)}^2\ge 0$ which is always true.