Is this inequality true: if $\alpha>0$ and $\beta>0$ , then $$|\alpha z_{1}+\beta z_2|\leq \max\{\alpha,\beta\}|z_1+z_2|,$$ for any $z_1,z_2\in\mathbb{C}$, such that $z_1\neq -z_2$.
Thanks!
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3 Answers
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[ EDIT ] The stronger statement can be proved that for any $\,\alpha \ne \beta\,$, there exist complex numbers $\,z_1,z_2\,$ such that the inequality does not hold.
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That's not true in general, consider for example $\alpha=1, \beta=2, z_1=1, z_2=-1+i\,$:
$$ |\alpha z_{1}+\beta z_2| = |-1+2i| = \sqrt{5} \;\;\color{red}{\gt}\;\; 2 = 2|i| = \max(\alpha,\beta)\,|z_1+z_2| $$
[ EDIT ] The stronger statement can be proved that for any $\,\alpha \ne \beta\,$, there exist complex numbers $\,z_1,z_2\,$ such that the inequality does not hold.
By symmetry, it can be assumed WLOG that $\,\alpha \lt \beta \,$, and by homogeneity that $\alpha=1, z_1=1$. With $\,z_2=z\,$ the proposed inequality then becomes:
$$\require{cancel} \begin{align} |1+\beta z| \le \beta |1+z| \;\;&\iff\;\; |1+\beta z|^2 \le \beta^2 |1+z|^2 \\ &\iff\;\;1 + \cancel{\beta^2|z|^2} + \beta(z+\bar z) \le \beta^2 \left(1+\cancel{|z|^2}+z+\bar z\right) \\ &\iff\;\; 1 - \beta^2 \le 2\beta(\beta-1)\operatorname{Re}(z) \\[5px] &\iff\;\; \operatorname{Re}(z) \ge \frac{1 - \beta^2}{2\beta(\beta-1)} = -\,\frac{\beta + 1}{2 \beta} \tag{*} \end{align} $$
Therefore the inequality does not hold true for any $\,z\,$ that does not satisfy the condition $\,(*)\,$.
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Suppose, without any loss of generality, that $\alpha > \beta$. Setting $\displaystyle\gamma:=\frac{\beta}{\alpha} < 1$, we obtain that \begin{align*} \alpha|z_{1}+z_{2}|\geq|\alpha z_{1}+\beta z_{2}| & \Leftrightarrow |z_{1} + z_{2}| \geq |z_{1} + \gamma z_{2}| \Leftrightarrow |z_{1} + z_{2}|^{2} - |z_{1} + \gamma z_{2}|^{2} \geq 0 \Leftrightarrow\\\\ |z_{1} + z_{2}|^{2} - \left|z_{1} + \gamma z_{2}\right|^{2} & = |z_{1}|^{2} + 2\mathrm{Re}(z_{1}\overline{z}_{2}) + |z_{2}|^{2} - |z_{1}|^{2} - 2\gamma\mathrm{Re}(z_{1}\overline{z}_{2}) -\gamma^{2}|z_{2}|^{2}\\\\ & = (1-\gamma^{2})|z_{2}|^{2} + 2(1-\gamma)\mathrm{Re}(z_{1}\overline{z}_{2}) \geq 0\\\\ & \Leftrightarrow (1+\gamma)|z_{2}|^{2} + 2\mathrm{Re}(z_{1}\overline{z}_{2}) \geq 0 \end{align*}
Hence $\max\{\alpha,\beta\}|z_{1}+z_{2}|\geq|\alpha z_{1}+\beta z_{2}|$ if, and only if, $\mathrm{Re}(z_{1}\overline{z}_{2})\geq -(1+\gamma)|z_{2}|^{2}/2$.
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YOUR inequality is not true!
Suppose $\alpha<\beta$, then $0<\frac{\alpha}{\beta}<1$. Take $z_1\neq 0$, your inequality is equivalent to $$\left|\frac{\alpha}{\beta} z_{1}+ z_2\right|\leq |z_1+z_2|,$$ that is to say $$\left|\frac{\alpha}{\beta} + \frac{z_2}{z_{1}}\right|\leq \left|1+\frac{z_2}{z_{1}}\right|.$$ So if you take $z_2$ such that $\frac{z_2}{z_{1}}=-2$, you will get $$\left|\frac{\alpha}{\beta} -2\right|\leq 1,$$ but this implies $$\frac{\alpha}{\beta}\geq1,$$ it is a contradiction.
