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.

By the AM-GM Inequality,
$$c|z_1|^2+\frac{1}{c}|z_2|^2\geq 2|z_1||z_2|\,.$$
Thus, $$(1+c)|z_1|^2+\left(1+\frac{1}{c}\right)|z_2|^2\geq \big(|z_1|+|z_2|\big)^2\geq |z_1+z_2|^2\,,$$
where the last inequality follows from the Triangle Inequality.  Note that the inequality $$(1+c)|z_1|^2+\left(1+\frac{1}{c}\right)|z_2|^2 \geq |z_1+z_2|^2$$ is an equality if and only if $z_2=cz_1$.

We also have
$$|z_1+z_2|^2\geq (1-c)|z_1|^2+\left(1-\frac1c\right)|z_2|^2\,.$$
The inequality above becomes an equality iff $z_2=-cz_1$.
