a^2+b^2+c^2+abc=4 and its consequences
by Sayan, Nov 11, 2012, 3:25 AM
If 
are non-negative reals such that 
, then we have the following nice inequalities:
Proof:![$4=abc+a^2+b^2+c^2 \ge abc+3\sqrt[3]{a^2b^2c^2}$](//latex.artofproblemsolving.com/2/2/6/2261e70575f57dc71a227b3c93050872bbef2c42.png)
Let 
then
![\[x^3+3x^2 \le 4 \implies (x-1)(x+2)^2 \le 0 \implies x \le 1 \implies abc \le 1\]](//latex.artofproblemsolving.com/3/6/8/36815d4e890b33e44a464a655981020db06f78c4.png)
Proof(mudok): Two of
have the same sign. Assume that 
. If 
then it is done. If 
then 
Contradiction. So 
Proof:
![\[abc \le 1 \implies 4-(a^2+b^2+c^2) \le 1 \implies a^2+b^2+c^2 \ge 3\]](//latex.artofproblemsolving.com/d/4/f/d4f7a35a4bb0ddc1ef30621bcf95ec2aa2254966.png)
(USAMO 2001): 
(mudok): 
(own) Refinement of 
: ![$a+b+c \ge ab+bc+ca+\frac{1}{5}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$](//latex.artofproblemsolving.com/e/5/1/e517161b8981d686b41e10583781eb18b012e360.png)
(own) Refinement of 
: ![$ab+bc+ca+\frac{1}{4}\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \le 3$](//latex.artofproblemsolving.com/2/0/b/20b2d5e21258cb7b2ac42ca9162b6cd9a4d7a7b0.png)
(mudok) ![$a+b+c \ge ab+bc+ca+\frac{1}{\sqrt{24}}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$](//latex.artofproblemsolving.com/0/4/e/04e449e69ecbd58f07e8357b0158f0bb6d492f57.png)
are non-negative reals such that 
, then we have the following nice inequalities:
Proof:
![$4=abc+a^2+b^2+c^2 \ge abc+3\sqrt[3]{a^2b^2c^2}$](http://latex.artofproblemsolving.com/2/2/6/2261e70575f57dc71a227b3c93050872bbef2c42.png)
Let 
then![\[x^3+3x^2 \le 4 \implies (x-1)(x+2)^2 \le 0 \implies x \le 1 \implies abc \le 1\]](http://latex.artofproblemsolving.com/3/6/8/36815d4e890b33e44a464a655981020db06f78c4.png)
Proof(mudok): Two of
have the same sign. Assume that 
. If 
then it is done. If 
then 
Contradiction. So 
Proof:
![\[abc \le 1 \implies 4-(a^2+b^2+c^2) \le 1 \implies a^2+b^2+c^2 \ge 3\]](http://latex.artofproblemsolving.com/d/4/f/d4f7a35a4bb0ddc1ef30621bcf95ec2aa2254966.png)
(USAMO 2001): 
(mudok): 
(own) Refinement of 
: ![$a+b+c \ge ab+bc+ca+\frac{1}{5}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$](http://latex.artofproblemsolving.com/e/5/1/e517161b8981d686b41e10583781eb18b012e360.png)
(own) Refinement of 
: ![$ab+bc+ca+\frac{1}{4}\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \le 3$](http://latex.artofproblemsolving.com/2/0/b/20b2d5e21258cb7b2ac42ca9162b6cd9a4d7a7b0.png)
(mudok) ![$a+b+c \ge ab+bc+ca+\frac{1}{\sqrt{24}}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$](http://latex.artofproblemsolving.com/0/4/e/04e449e69ecbd58f07e8357b0158f0bb6d492f57.png)
This post has been edited 2 times. Last edited by Sayan, Dec 15, 2012, 1:30 PM
![\[(1+a)(1+b)(1+c)\ge (a+b)(b+c)(c+a)\]](http://latex.artofproblemsolving.com/2/3/3/23343d1966dd024fec35d6a5aaffe9e443381c0e.png)
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