<s>[INST] 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.
 [/INST] 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).
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