Prove that $|z_{1}+z_{2}|^2\leq (1+c)|z_{1}|^2+\bigg(1+\frac{1}{c}\bigg)|z_{2}|^2$ 
If $z_{1},z_{2}$ are two complex numbers and $c>0.$ Then prove that
$\displaystyle |z_{1}+z_{2}|^2\leq (1+c)|z_{1}|^2+\bigg(1+\frac{1}{c}\bigg)|z_{2}|^2$

Try: put $z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}.$ Then
from left side
$$(x_{1}+x_{2})^2+(y_{1}+y_{2})^2=x^2_{1}+x^2_{2}+2x_{1}x_{2}+y^2_{1}+y^2_{2}+2y_{1}y_{2}$$
Could some help me how to solve it further, Thanks in Advance.

Let $a=|z_1|,b=|z_2|$. Then $2ab=2(a\sqrt c)  \frac b {\sqrt c} \leq ca^{2}+\frac {b^{2}} c$ (because $2ts \leq t^{2}+s^{2}$ for any two real numbers $t$ and $s$). Hence $|z_1+z_2|^{2} \leq (a+b)^{2}=a^{2}+b^{2}+2ab\leq (1+c)a^{2}+(1+\frac 1 c )b^{2}$.
