# Minkowski Inequality for $p \le 1$

I've been trying to prove the concavity of a particular function which I reduced to proving the reverse Minkowski Inequality for $p \le 1$, $p \ne 0$ for arguments in $\mathbb{R}^{n}_{+}$. That is,

$$\|x \|_p + \|y\|_p \le \|x+y\|_p$$

I've found a proof for the case of $0 < p \le 1$, but the case when $p < 0$ still evades me. All proofs of Minkowski's Inequality (in the proper direction) usually rely on Hölder's Inequality, which in turn relies on Young's Inequality. However, Young's does not apply for exponents below $0$, and I am rather jammed up finding another way. Can anyone offer a little direction?

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Isn't that false, at the very least for $p=1$? Together with the usual Minkowski inequality you deduce that $$\| x+y \| _p = |\ x\|_p + \| y \|_p ,$$ which is false generally. – Ragib Zaman Oct 17 '11 at 9:12
Quite right you are if either $x,y < 0$. I've amended it to only consider values with elements greater than or equal to 0. – duckworthd Oct 17 '11 at 21:28

Hölder is reversed when $p<0$. The proof is a simple change of variables. See Bullen's book "Handbook of Means and Their Inequalities" p.202-3 for details. Hence Minkowski is also reversed.
Using the concavity of the function $f$ defined by $f(x)=x^p$, $0<p<1$, we get that for $t\in (0,1)$, $$(x_k+y_k)^p=\bigg(t\dfrac{x_k}{t}+(1-t)\dfrac{y_k}{1-t}\bigg)^p\geq t\dfrac{x_k^p}{t^p}+(1-t)\dfrac{y_k^p}{(1-t)^p}.$$ Taking the sum, we get that $$||x+y||_p^p\geq t\dfrac{||x||_p^p}{t^p}+(1-t)\dfrac{||y||_p^p}{(1-t)^p}.$$ By choosing $t=\dfrac{||x||_p}{||x||_p+||y||_p}$ and $1-t=\dfrac{||y||_p}{||x||_p+||y||_p}$, we get that