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I need to prove: for non-negative functions $f,g\in L^p[0,1]$, $||f+g||_p\geq||f||_p+||g||_p$ for $0<p<1$.

For $1\leq p<\infty$, the inequality is reversed and the proof is like: The cases $||f||_p=0$ or $||g||_p=0$ are trivial. So we assume $||f||_p=\alpha\neq0$ and $||g||_p=\beta\neq0$, then there are functions $f_0,g_0\in L^p$ such that $|f|=\alpha f_0,|g|=\beta g_0$ (then $||f_0||_p=||g_0||_p=1$), and defining $\lambda=\frac{\alpha}{\alpha+\beta}$, \begin{eqnarray} |f(x)+g(x)|^p&\leq&(|f(x)|+|g(x)|)^p=[\alpha f_0+\beta g_0]^p~~~~~~~~~~(*)\\ &=&(\alpha+\beta)^p[\lambda f_0(x)+(1-\lambda)g_0(x)]^p\\ &\leq&(\alpha+\beta)^p[\lambda f_0(x)^p+(1-\lambda)g_0(x)^p]~~~~~~~(**) \end{eqnarray} by convexity of the function $\phi(t)=t^p$ for $1\leq p<\infty$. The rest follows by taking integration of both sides over $[0,1]$.

Now it says in the hint that for the result I want to prove for $0<p<1$, we need only to use the concavity of the function $\phi(t)=t^p$ for $0<p<1$. But that only reverses the inequality for $(**)$ and not $(*)$.

Then how can I prove it?

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The result you want isn't true without some additional hypothesis - maybe $f,g \ge 0$. Otherwise $f = -g$ is a counterexample. But my suggested added hypothesis makes your first inequality an equality, and so you don't have to worry about it.

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  • $\begingroup$ extremely sorry, forgot to mention. $f$ and $g$ are non-negative functions. I edited my question. $\endgroup$ – Abishanka Saha Dec 29 '13 at 5:46
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    $\begingroup$ But then you have $|f(x) + g(x)|^p = (|f(x)|+|g(x)|)^p$ and your problem is solved. $\endgroup$ – Stephen Montgomery-Smith Dec 29 '13 at 5:56
  • $\begingroup$ Oh! How that escaped me!! That was very helpful!! Thank you! $\endgroup$ – Abishanka Saha Dec 29 '13 at 5:58
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    $\begingroup$ You can also mimic the proof of Minkowski that goes via Hoelder's inequality. But here the conjugate index is negative. Again the inequality is reversed, and non-negativity is required. $\endgroup$ – Stephen Montgomery-Smith Dec 29 '13 at 6:41

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