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Given $a,b,c\geq 0$, then this following is true $$\min(a,b+c)\leq \min(a,b)+ \min(a,c)$$

I tried to prove it by using some cases, i.e.
Case 1: if $\,\,a\leq b \leq c\,\,$ and $\,\,a\leq c \leq b,\,\,$ it's clear that \begin{align*} \min(a,b+c)&\leq \min(a,b)+\min(a,c)\\ a &\leq a+a=2a \end{align*} Case 2: if $\,\,b\leq a\leq c\,\,$ and $\,\,c\leq a \leq b,\,\,$ we have
\begin{align*} \min(a,b+c)&\leq \min(a,b)+\min(a,c)\\ a &\leq b+a \end{align*} and \begin{align*} \min(a,b+c)&\leq \min(a,b)+\min(a,c)\\ a &\leq a+c \end{align*} Case 3: $\,\,c\leq b\leq a\,\,$ and $\,\,b\leq c \leq a,\,\,$
I am not sure about this case, because although $a$ is the biggest from $b$ and $c$, it can satisfies $a\leq b+c$ or $b+c\leq a$. Is the third case must be done with two cases again? i.e for $a\leq b+c$ and $b+c\leq a$?

or there is another way to prove $\min(a,b+c)\leq \min(a,b)+ \min(a,c)$ for $a,b,c\geq 0$?

Any help will be appreciated. Thanks.

Neat Math
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user136524
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3 Answers3

3

For your case 3: $\min\{a,b\} + \min\{a,c\} = b+c$. Since $\min\{a,b+c\}\leq b+c$ (min is always less than both numbers) you get the inequality.

Rida
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1

Alternative proof:

Case 1: $a\ge b+c.$ Then $a\ge b, a\ge c$. $$\min(a,b+c)=b+c =\min(a,b)+ c=\min(a,b)+\min(a,c).$$

Case 2: $a \le b+c.$ Then $\exists x, y \ge 0, a=x+y$, such that $x\le b, y \le c$. (For example $x=\frac{ab}{b+c}, y=\frac{ac}{b+c}.)$

Then $$ x \le a, x \le b \implies x \le \min(a,b)$$ $$ y \le a, y \le c \implies y \le \min(a,c)$$ Therefore $$ a=x+y \le \min(a,b)+\min(a,c). \blacksquare $$

Neat Math
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1

There's another way. Consider the function $f(a)=\min(a,b)+\min(a,c)-\min(a,b+c)$. It's piece-wise linear, so to prove it's non-negative, we only need to check its value at each endpoint of the intervals (including $+\infty$):

$f(0)=0+0-0=0;$

$f(b)=b+\min(b,c)-\min(b,b+c)=\min(b,c) \ge 0.$

By symmetry $f(c)=\min(b,c)$.

$f(b+c)=b+c-(b+c)=0.$

And lastly, when $a\to \infty, f(a)=b+c-(b+c)=0.$

Some other problems can be solved this way as well:

Show $\frac{1}{\lvert a - b \rvert} < \frac{2}{\lvert a \rvert}$ : This one you just check the function $f(b)=2|a-b|-|a|$ at $b=\frac{|a|}{2} - \varepsilon, -\frac{|a|}{2}+ \varepsilon$, with $\varepsilon>0$ arbitrarily small.

Prove that $|a|\leq \max\{|b|,|c|\}$ if $b\leq a \leq c$ : This one you check $a=0, b, c$ (even though $0$ may be outside of $[b,c]$).

Even equations: Why does the equality $\max\{a,\min\{b,c\}\}=\min\{\max\{a,b\},\max\{a,c\}\}$ hold? where you check $a=\min(b,c), b,c, -\infty, +\infty$.

Or even this one Reverse Triangle Inequality Proof albeit not the best way.

Neat Math
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