If $a\le b$ and $l,m\ge 1$, then $|l+e^{i\gamma}|(a+mb)\leq (l+m)|a+e^{i\gamma}b|$

Question

Let $a, b$ be any two positive real numbers such that $a\geq lb$ where $l\geq 1.$ Suppose $\gamma $ is any real such that $0\leq \gamma\leq 2\pi.$ Is it true that $$|l+e^{i\gamma}|(a+mb)\leq (l+m)|a+e^{i\gamma}b|$$ for any $m\geq 1?$

Answer 1

This is equivalent to: $$\frac{a+mb}{l+m}\leq\left|\frac{a+e^{i\gamma}b}{l+e^{i\gamma}}\right|$$ or to: $$b+\frac{a-bl}{l+m}\leq \left|b+\frac{a-bl}{l+e^{i\gamma}}\right| $$ that follows from: $$b+\frac{a-bl}{l+m}\leq b+\frac{a-bl}{l+1}\leq\left|b+\frac{a-bl}{l+e^{i\gamma}}\right|. $$