I thought about this while studying about $L^p$ spaces, as the standard triangle inequality does not hold for $0<p\leq 1$. But we have the variant
$$||f+g||_{L^p}^p \leq ||f||_{L^p}^p+||g||_{L^p}^p .$$
To simplify the question, what would be the best (or general) way of proving the type of inequalities that involve $|a+b|^p$ and $|a|^p + |b|^p$?
There is the standard trick
$$|a+b|^p \leq (2\max(|a|,|b|))^p = 2^p \max(|a|^p, |b|^p) \leq 2^p (|a|^p+ |b|^p),$$
but there is the factor of $2^p$. One can also normalize one side of inequality, or use convexity argument, etc.
For example, $|a+b|^p \leq |a|^p + |b|^p$ for $0<p\leq 1$ , I had
$$1 = \frac{|a|}{|a|+|b|} +\frac{|b|}{|a|+|b|}\leq \left(\frac{|a|}{|a|+|b|} \right)^p +\left(\frac{|b|}{|a|+|b|} \right)^p.$$
Edit: Thank you for the replies! I will add 1 remark:
For $1\leq p<\infty$, we have
$$\left( \sum |a_i+b_i|^p \right)^{\frac{1}{p}} \leq \left( \sum (|a_i|+|b_i|)^p \right)^{\frac{1}{p}}\leq \left( \sum |a_i|^p \right)^{\frac{1}{p}}+\left( \sum |b_i|^p \right)^{\frac{1}{p}},$$
$$\sum (|a_i|+|b_i|)^p \geq \sum |a_i|^p +\sum |b_i|^p .$$
For $0 < p \leq 1$, we have
$$\left( \sum (|a_i|+|b_i|)^p \right)^{\frac{1}{p}} \geq \left( \sum |a_i|^p \right)^{\frac{1}{p}}+\left( \sum |b_i|^p \right)^{\frac{1}{p}},$$
$$\sum |a_i+b_i|^p \leq \sum (|a_i|+|b_i|)^p\leq\sum |a_i|^p +\sum |b_i|^p .$$
Is there a intuitive or visual reasoning that the inequalities behave his way?
Best Answer
If you have the inequality you found:
\begin{align} 1=\frac{|a|}{|a|+|b|}+\frac{|b|}{|a|+|b|}& \le\left(\frac{|a|}{|a|+|b|}\right)^p+\left(\frac{|b|}{|a|+|b|}\right)^p=\frac{|a|^p+|b|^p}{(|a|+|b|)^p}\\ \implies(|a|+|b|)^p & \le|a|^p+|b|^p \end{align}
Also, $(|a+b|)^p\le(|a|+|b|)^p\le|a|^p+|b|^p$, so you get the desired result.