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.
 A: You are on the right track. Following your approach the inequality becomes
$$x^2_{1}+x^2_{2}+2x_{1}x_{2}+y^2_{1}+y^2_{2}+2y_{1}y_{2}\leq (1+c)(x_1^2+y_1^2)+\bigg(1+\frac{1}{c}\bigg)(x_2^2+y_2^2)$$
that is
$$2x_{1}x_{2}+2y_{1}y_{2}\leq c(x_1^2+y_1^2)+\frac{1}{c}(x_2^2+y_2^2).$$
Is it true that $2uv\leq cu^2+\frac{v^2}{c}$? (Hint. use AM-GM inequality).
A: 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 $z_2=-cz_1$.
A: 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 $|z_1+z_2|^{2} \leq (a+b)^{2}=a^{2}+b^{2}+2ab\leq (1+c)a^{2}+(1+\frac 1 c )b^{2}$.
