$a,b,c \geq 0$ and $a^2+b^2+c^2+abc=4$ prove that $ab+bc+ac-abc \leq 2$ can any one help me with this problem,I believe Dirichlet's theorem is the key for this sorry for making mistake over and over again,but i'm certain that the inequality is true now.
Asked
Active
Viewed 1,552 times
-1
-
1$a+b+bc+ac-abc\le 2$ is correct? Or $ab+bc+ca-abc\le 2$? – mathlove Jul 31 '14 at 12:37
-
$a+b+bc+ac−ab\leq2$ is true, – Aerrozard Jul 31 '14 at 12:41
-
OK. That's all fine, then. – mathlove Jul 31 '14 at 12:43
-
@user167099: You mean that $a+b+bc+ac-abc\le 2$ is true, right?(typo at $-ab$) – mathlove Jul 31 '14 at 12:54
-
yep.my bad :Dbut the part $ a+b$ is true – Aerrozard Jul 31 '14 at 12:57
-
Double check your Q. Counter e.g. $(a, b, c) \approx (1.97852, 0.1875, 0.105589)$. – Macavity Jul 31 '14 at 13:18
3 Answers
1
Note $a^2+b^2+c^2+abc=1 \implies 0\le a,b,c \le 1$. Hence $$(1-a)(1-b)c\ge 0 \implies c\ge bc+ca-abc$$
So now it is enough to prove $a+b+c \le 2$, which is obvious from AM-QM $$\frac{a+b+c}3 \le \sqrt{\frac{a^2+b^2+c^2}3}\le \sqrt{\frac13}$$
Macavity
- 46,381
-
-
2@user167099 Why don't you then double check the question? Because with $4$, the inequality in your problem will not hold true. – Macavity Jul 31 '14 at 13:17
0
I assume you want $ab+bc+ca-abc\leq2$. If this is not the case, let me know. This is not the best inequality you could get, but it's true.
Since $(a-b)^2\geq0$, we have $a^2+b^2\geq2ab$, and similarly $b^2+c^2\geq2bc$ and $c^2+a^2\geq2ca$. Summing up the three inequalities gives $$ 2(a^2+b^2+c^2)\geq2(ab+bc+ca). $$ If you combine this with $a^2+b^2+c^2+abc=1$, you get $ab+bc+ca+abc\leq1$.
But now $2abc\geq0$ so you can subtract it from the left side. Similarly you can add one to the right side. This leads to $ab+bc+ca-abc\leq2$.
Joonas Ilmavirta
- 25,809
0
For your last revision of the question (you'll need to change the title too)...
Note among any three numbers $a,b,c$, some two will be either $\le 1$ or $\ge 1$. WLOG let these be $a,b$. So $(1-a)(1-b)c \ge 0 \implies bc + ca \le c+abc \implies ab+bc+ca -abc\le ab+c$. Now you need to show $ab \le 2-c$. For that, note:
$$4-c^2 = a^2+b^2+abc \ge 2ab+abc = ab(2+c)$$
P.S. You may want to check http://www.artofproblemsolving.com/Forum/blog.php?b=78664 for more variants, but don't keep changing the question!
Macavity
- 46,381
-
Note we have used Dirichlet / Pigeonhole in the above proof, I guess it can be done without it, but probably much more cumbersome. – Macavity Aug 01 '14 at 08:33