Dtyjlui

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?

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 $t = dfrac vu$.

It is also a consequence of Minkowski's inequality for $p=frac{3}{2}$. Let
$$
stackrel{rightarrow}{a}=(u,0),quadstackrel{rightarrow}{b}=(0,v).
$$ Then Minkowski's inequality says
$$
left(u^{frac{3}{2}}+v^{{frac{3}{2}}}right)^{frac{2}{3}}=|stackrel{rightarrow}{a}+stackrel{rightarrow}{b}|_{frac{3}{2}}le |stackrel{rightarrow}{a}|_{frac{3}{2}}+|stackrel{rightarrow}{b}|_{frac{3}{2}}=u+v.
$$

It's
$$u^3+3u^2v+3uv^2+v^3geq u^3+2sqrt{u^3v^3}+v^3$$ or
$$3u+3vgeq2sqrt{uv}$$ or
$$(sqrt{u}-sqrt{v})^2+2u+2vgeq0,$$ which is obvious.

Write $a= sqrt{u}$ and $b= sqrt{v}$, so you have to prove:

$$ (a^3+b^3)^2leq (a^2+b^2)^3$$
so $$ 2a^3b^3leq 3a^4b^2+3a^2b^4$$

so $$2ableq 3a^2+3b^2$$

But this is true since $$2ableq a^2+b^2leq 3a^2+3b^2$$

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

$$2sqrt{uv} le 3u+3v$$

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 $2sqrt{uv} le 3u+3v$ as well.