Convexity of $ (a+b)^{1/n}$

Question

How to prove that $(a+b)^{1/n} \le a^{1/n}+b^{1/n}$ by the convexity of $(a+b)^{1/n}$

Answer 1

Welcome to MSE: as a hint take $f(x)=\sqrt[n]{x+1}$ and prove for concavinty of $f(x)$ then use $$\sqrt[n]{x+1}\leq \sqrt[n]{x}+\sqrt[n]{1}$$ now it suffice to put $x=\frac ab$ and multiply both sides by $\sqrt[n]{b}$

extra hint: if you accept $f(x)=\sqrt[n]{x+1}$ is concave function,so $$\sqrt[n]{x+1}\leq \sqrt[n]{x}+\sqrt[n]{1}\\ \text{ put } x=\frac ab \to \\\sqrt[n]{\frac ab+1}\leq \sqrt[n]{\frac ab}+\sqrt[n]{1}\\\text{ multiply by} \sqrt[n]{b} \text{both sides} $$