$(x+y)^a \le x^a+y^a$ for $x,y \ge 0$ and $0

Question

Trying to prove $(x+y)^a \le x^a+y^a$ for $x,y \ge 0$ and $0<a<1$. I found one way, using derivatives: The inequality is true for x=y=0 and the partial derivatives by x or y follow the same inequality.

My question is if this is a special case of something deeper than this game of derivatives.

score 4 · Accepted Answer · answered Jul 03 '11 at 14:55

4

It may be worth noting that, writing $y=cx$, for $c \geq 0$, the original inequality is equivalent to $$ (1+c)^a \leq 1 + c^a, $$ where $0 < a <1$, which is a more elegant inequality to handle.

answered Jul 03 '11 at 14:55

Shai Covo

24,077

Shai: This last statement of yours is highly debatable. Ironically, I remember reading somewhere that the first thing to do with inequalities similar to the one you suggest was to make them homogenous... hence going back to the OP's formulation. – Did Jul 03 '11 at 18:34

score 2 · Answer 2 · answered Jul 03 '11 at 14:39

2

In any case, $u(x)+u(y)\le u(x+y)+u(0)$ as soon as $u'$ is nonincreasing as soon as $u''\le0$. (No partial derivative is needed here.)

answered Jul 03 '11 at 14:39

Did

279,727

$(x+y)^a \le x^a+y^a$ for $x,y \ge 0$ and $0

2 Answers2

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