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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).

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This does not require any of these. Who told you so? – 1015 Apr 10 '13 at 12:49
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 $b\geq 1$,

$$ \frac{x}{a}\leq ax\quad\mbox{and}\quad \frac{y}{b}\leq by\quad\Rightarrow \quad\frac{x+y}{a+b}\leq \max\{ax,by\}\qquad\forall x\ge 0,y\ge 0,a\geq 1,b\geq 1. $$

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Great proof. I see my (a)-(c) ideas were wrongs. I made no typos though. Thanks. – Nicolas Essis-Breton Apr 10 '13 at 13:36
You're welcome, Nicolas. These were not bad ideas. – 1015 Apr 10 '13 at 13:38

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