# Prove that at least two of these inequalities are true: $|a-b|\le2$, $|b-c|\le2$, $|c-a|\le2$.

It's given that: $$\begin{cases}a,b,c>0\\a+b+c\le4\\ab+bc+ac\ge4\end{cases}$$

Prove without using calculus that it's true that at least two of these are correct inequalities:$$\begin{cases}|a-b|\le2\\|b-c|\le2\\|c-a|\le2\end{cases}$$

If you think about it, we may as well prove that at least two of these are correct:$$\begin{cases}(a-b)^2\le4\\(b-c)^2\le4\\(c-a)^2\le4\end{cases}$$

We could square one of the inequalities (since both sides are positive):$$(a+b+c)^2\le16\Rightarrow a^2+b^2+c^2+2(ab+bc+ac)\le16$$

$$a^2+b^2+c^2+8\le a^2+b^2+c^2+2(ab+bc+ac)\le16\Rightarrow a^2+b^2+c^2\le 8$$

So now we know that: $$\begin{cases}a,b,c>0\\a+b+c\le4\\a^2+b^2+c^2\le8\\ab+bc+ac\ge4\end{cases}$$

Does anyone see how to solve this? Thanks.

• It's best not to use \mid except in contexts like $3\mid 12$ and $\{x\in\Bbb Q\mid 2x+3\in\Bbb Z\},$ as it messes up the spacing. Rather, simply use the vertical bar on your keyboard. Commented Dec 23, 2013 at 15:49
• a,b,c are real or integers? Commented Dec 23, 2013 at 15:49
• @Chris They are positive real numbers. I thought it would be clear, sorry for not pointing it out. Commented Dec 23, 2013 at 15:50

## 2 Answers

Suppose two of these are false. Then $$8<(a-b)^2+(a-c)^2+(b-c)^2=2(a^2+b^2+c^2)-2(ab+ac+bc)\le 16-8$$ i.e. $8<8$, a contradiction.

• I think you miss the case when only one of them is correct. Commented Dec 23, 2013 at 15:54
• @Shuchang Should be fixed now. Commented Dec 23, 2013 at 15:55

It is easy! first, we know that $a^2+b^2+c^2<=8$, therefore

$2a^2+2b^2+2c^2<=16$ ............. ine1

and we know $ab+ba+ca>=4$,so

$-2ab-2bc-2ca<=-8$ ......... ine2

so ine1+ine2, we get

$(a-b)^2+(b-c)^2+(a-c)^2<=8$

so ,it is easy to explain your inequalities:

$\begin{cases}(a-b)^2\le4\\(b-c)^2\le4\\(c-a)^2\le4\end{cases}$

at least two of there are right.

• You meant $(a-b)^2+(b-c)^2+(a-c)^2$. Commented Dec 23, 2013 at 16:07
• @mathh yes, edited. Commented Dec 23, 2013 at 22:42