How to show $(u^{3/2}+v^{3/2})^{2/3} \leq u+v$ where $u,v \geq 0$? In Convex Optimization book written by Prof. Boyd on page 499 section 9.6.2 says we have 
$$(u^{3/2}+v^{3/2})^{2/3} \leq u+v$$
where $u,v \geq 0$. How can we show that?
 A: In general if $p > 1$ and $u,v \ge 0$ then $u^p + v^p \le (u+v)^p$. This is evident if $u = 0$, and if $u > 0$ you can divide by $u^p$ to arrive at the equivalent inequality
$$1 + \left( \frac vu \right)^p \le \left( 1 + \frac vu \right)^p.$$
The function $\phi(t) = (1+t)^p - t^p - 1$, $t \ge 0$, satisfies $\phi'(t) = p(t+1)^{p-1} - pt^{p-1} > 0$ for $t > 0$ because $p-1 > 0$. Thus $\phi$ is increasing on $[0,\infty)$ so that $\phi(t) \ge \phi(0) = 0$. That is,
$$1 + t^p \le (1+t)^p$$ for all $t \ge 0$ and in particular for $t = \dfrac vu$.
A: It is also a consequence of Minkowski's inequality for $p=\frac{3}{2}$. Let
$$
\stackrel{\rightarrow}{a}=(u,0),\quad\stackrel{\rightarrow}{b}=(0,v).
$$ Then Minkowski's inequality says
$$
\left(u^{\frac{3}{2}}+v^{{\frac{3}{2}}}\right)^{\frac{2}{3}}=\|\stackrel{\rightarrow}{a}+\stackrel{\rightarrow}{b}\|_{\frac{3}{2}}\le \|\stackrel{\rightarrow}{a}\|_{\frac{3}{2}}+\|\stackrel{\rightarrow}{b}\|_{\frac{3}{2}}=u+v.
$$
A: It's
$$u^3+3u^2v+3uv^2+v^3\geq u^3+2\sqrt{u^3v^3}+v^3$$ or
$$3u+3v\geq2\sqrt{uv}$$ or
$$(\sqrt{u}-\sqrt{v})^2+2u+2v\geq0,$$ which is obvious.
A: Write $a= \sqrt{u}$ and $b= \sqrt{v}$, so you have to prove:
$$ (a^3+b^3)^2\leq (a^2+b^2)^3$$
so $$ 2a^3b^3\leq 3a^4b^2+3a^2b^4$$
so $$2ab\leq 3a^2+3b^2$$
But this is true since $$2ab\leq a^2+b^2\leq 3a^2+3b^2$$
A: I would raise each side to the $3^{rd}$ power and expand:
$$u^3+2(uv)^{\frac 32} + v^3 \le u^3 + 3u^2v + 3uv^2 + v^3$$ 
Which simplifies to 
$$2(uv)^{\frac 32} \le 3uv(u+v)$$
Which is 
$$2\sqrt{uv} \le 3u+3v$$
And if you are familiar with the famous inequality $a^2+b^2 \ge 2ab$ (which is gotten from $(a-b)^2 \ge 0$) then you see that $2\sqrt{uv} \le 3u+3v$ as well.
