$L^p$ norm and triangle inequality 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?
 A: 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.
A: Since $|a+b| \le |a| + |b|$ you have also that $|a+b|^p \le (|a| + |b|)^p$. It suffices to show that $(|a|+|b|)^p \le |a|^p + |b|^p$. The method you propose (completed by ellya in a different answer) is very short and elegant, but the solution I will give is perhaps a bit more flexible. 
You can start by taking $|a| = 1$ and proving that $(1+t)^p \le 1 + t^p$ for all $t \ge 0$. The standard method is the first derivative test. Define $\phi(t) = (1+t)^p - 1 - t^p$. Then $\phi'(t) = p(1+t)^{p-1} - p t^{p-1}$. Since $p < 1$, $p-1$ is negative. This means $t^{p-1} > (1+t)^{p-1}$ for all $t > 0$. It follows that $\phi'(t) < 0$ for all $t > 0$ (here we use $p > 0$) and consequently $\phi$ is decreasing. Since $\phi(0) = 0$, you conclude $\phi(t) \le 0$ for all $t \ge 0$. That is, $(1+t)^p \le 1 + t^p$.
Now for the general case: if $|a| \not= 0$ you have 
$$\left( 1 + \frac{|b|}{|a|} \right)^p \le 1 + \left(\frac{|b|}{|a|}\right)^p$$ so that you can multiply both sides by $|a|^p$ to get the inequality. If $|a| = 0$ the inequality is trivial.
