IIT JEE , AIEEE Forum - Mathematics , Algebra

anchitsaini

Let a,b,c be non negative real numbers such that

a² + b²+ c² + abc = 4

Prove that --

0 <= ab + bc + ca - abc <= 2

(there are more than one solutions)

sandeepramesh

LHS is done bcos abc < 4 and ofc by AM GM ab + bc + ca > = 3 (abc)^2/3 and 3 (abc)^2/3 > abc

sandeepramesh

n u could have said positive nos bcos clearly the trivial case isnt valid :D

sandeepramesh

2nd part:

a^2 + b^2 + c^2 > ab + bc +ca

implies ab+bc+ca - abc < = 4 - 2abc

But a^2+ b^2 + c^a > abc for abc < 4?

Hence ab+bc+ca<=2

hsbhatt

AM-GM gives abc

1

Now a²+b²+c² = 4-abc

ab+bc+ca

Hence 4-2abc

ab+bc+ca-abc

4-2abc

2

Hence ab+bc+ca

2

beaten to the finish

sandeepramesh

am i right then?

hsbhatt

didnt want 2 point out, but looks highly suspect!

sandeepramesh

what does that mean sirji?

hsbhatt

your last two lines of the proof are making me dizzy. Cud u write it down more elaborately for dummies like yours truly

sandeepramesh

well i kinda said in a genl way not wanting to find out the answer meself :P

i knew for sure that abc <4 and also a^2 + b^2 + c^2 > abc and so i didnt bother proving that but just stated it as a fact to prove later but u had already done that :D

And ofc the last line is the same as yours with using the eqn a^2 + b^2 + c^2 + abc =4

hsbhatt

bitte langsam!

See a²+b²+c²>abc

Hence, a²+b²+c²+abc = 4>2abc

So ab+bc+ca = 4-2abc<0.

Was macht du Herr Rohit!

just realisedmy proof is nonsense

I get the feeling that the problem is flawed

sandeepramesh

what are you trying to do there? and i dont understand hindi :D

no i said ab+bc+ca<=a^2+b^2+c^2<= 4 - abc

abc < 2 Hence the result follows

anchitsaini

i'll post the soln (solutions!!)
tomorrow!!

till then keep posting!

hsbhatt

I want to approach this like a police interrogator.

Does the name Zuming Feng mean anything to you?

sandeepramesh

yes co author with Titu Andreescu in a combinatorics book

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