Take the 2-minute tour ×  Let $z=x+y$ with $x,y\ge0$ and
$a,b>1$. Show that $$ \max\{ax,by\} \ge \frac{1}{a+b}z. \tag{1}
$$  This requires either the use of: (a) the convex function
$f(x)=\frac{1}{x}$, (b) the harmonic-arithmetic mean inequality,
or, (c) the Cauchy-Schwarz inequality.  I don't see how to link
(1) with any of (a)-(c).  share|improve this question This does
not require any of these. Who told you so? –  1015 Apr 10 '13 at
12:49  1 Answer 1  up vote 2 down vote accepted  First, note
that for every $x',y'\in\mathbb{R}$ and every $a>0,b>0$, we have
$$ \frac{ax'+by'}{a+b}\leq
\frac{a\max\{x',y'\}+b\max\{x',y'\}}{a+b}=\max\{x',y'\}. $$ In
convex terms, this is just because the lhs belongs to the
interval whose bounds are $x'$ and $y'$.  With $x'=x/a$ and
$y'=y/b$, we get  $$ \frac{x+y}{a+b}\leq
\max\left\{\frac{x}{a},\frac{y}{b}\right\}\qquad \forall
x,y\in\mathbb{R}\;\forall a>0,b>0. $$  Now for $x\geq 0$ and
$y\geq 0$, and $a\geq 1$ and $b\geq 1$,  $$ \frac{x}{a}\leq
ax\quad\mbox{and}\quad \frac{y}