Let $a>0,k>1$. Prove: $|a+z|\ge \frac{a+|z|}{k}$ when $|\arg(z)|\le 2\arccos\frac{1}{k}$ Let $a>0,k>1$. Prove: $|a+z|\ge \frac{a+|z|}{k}$ when $|\arg(z)|\le 2\arccos\frac{1}{k}$
Could someone give a hint how to prove this?

I like this your problem, following is my answer.
Let $z=re^{it}$, then we have $$|t|\le 2\arccos{\frac{1}{k}}\Longrightarrow k\cos{\dfrac{t}{2}}\ge 1$$
$$\Longleftrightarrow k|a+z|\ge |a|+|z|\Longleftrightarrow k^2|a+re^{it}|^2\ge (a+r)^2$$
$$\Longleftrightarrow k^2(a+re^{it})(a+re^{-it})\ge a^2+r^2+2ar$$
$$\Longleftrightarrow k^2(a^2+r^2+2ar\cos{t})\ge a^2+r^2+2ar$$
$$\Longleftrightarrow (k^2-1)(a^2+r^2)+2ar(k^2\cos{t}-1)\ge 0$$
since
$$a^2+r^2\ge 2ar,k^2-1\ge 0$$
so we only prove
$$k^2-1+k^2\cos{t}-1\ge 0$$
$$\Longleftrightarrow k^2(1+\cos{t})-2\ge 0$$
$$
\Longleftrightarrow k^2\cos^2{\dfrac{t}{2}}\ge 1$$
it is clear
