$(x+y)^{p/2} \leq x^{p/2} + y^{p/2} (1

Question

Let $x, y \in (0, \infty)$ and $1<p<2.$

My Question: Is it true that, $$(x+y)^{p/2} \leq x^{p/2} + y^{p/2} ?$$

Answer 1

Define $f(x)=(x+1)^{\frac{p}{2}}-x^{\frac{p}{2}}-1$, $x>0$. Note that, $f'(x)=\frac{p}{2}\left((x+1)^{\frac{p-2}{2}}-x^{\frac{p-2}{2}}\right)\le0$, because $x+1\ge x$ and $\frac{p-2}{2}<0.$ So, for $x>0$, $f$ is decreasing, then $f(x)\le f(0)$ for $x>0$. Therefore, $$(x+1)^{\frac{p}{2}}\le x^{\frac{p}{2}}+1.$$ Replace $x$ for $\frac{x}{y}$, $x,y>0$, we obtain, $$\left(\frac{x}{y}+1\right)^{\frac{p}{2}}\le \left(\frac{x}{y}\right)^{\frac{p}{2}}+1.$$ Therefore, $$\left(x+y\right)^{\frac{p}{2}}\le x^{\frac{p}{2}}+y^{\frac{p}{2}}.$$

Answer 2

Divide both sides by $(x+y)^{p/2}$ to get $$1 \le (\frac{x}{x+y})^{p/2}+(\frac{y}{x+y})^{p/2}$$

Since $x,y>0$ then $t=\frac{x}{x+y}$ and $1-t=\frac{y}{x+y}$ are in the interwal $(0,1)$.

So let us consider the function $f(t)=t^{q}+(1- t)^q$ for $q\in (1/2, 1)$ and $t \in [0,1]$.

$f'(t)=qt^{q-1}-q(1-t)^{q-1}=q(t^{q-1}-(1-t)^{q-1})=0.$

Let me continue:

$f'(t)=0 \Leftrightarrow t^{q-1}-(1-t)^{q-1}=0 \Leftrightarrow t^{q-1}=(1-t)^{q-1} \Leftrightarrow t=1-t$. Therefore $t=1/2.$

For $t \in (0,1/2), t^{q-1}>(1-t)^{q-1} \Rightarrow f'(t)>0$.

For $t \in (1/2,1), t^{q-1}<(1-t)^{q-1} \Rightarrow f'(t)<0$.

Therefore the smallest value of $f(t)$ either $f(0)=1$ or $f(1)=1$.

Hence $f(t) \ge 1$ for $t \in [0,1]$. This completes the proof.