<s>[INST] 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.
 [/INST] 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. 
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