How to prove this?

Number Theory Level 3

$(abc)^4+abc(a^3c^2+b^3a^2+c^3b^2)\le K$

If $a,b,c>0$ and $a + b + c = 3$ ,

Find $K$

2 solutions

Department 8
Jul 15, 2015

We have $a+b+c=3$ We were given that $a, b, c$ should be non-zero positive integers which leaves us only choice $a=b=c=1$

It's not given that they are integers -_-

Raushan Sharma - 5 years, 4 months ago

I don't know when I posted this I will check on it later for the solution.

Department 8 - 5 years, 4 months ago

Good thing I assumed they were integers. :D

Guy Alves - 4 years, 6 months ago

Pavan Kartik
Jun 13, 2020

apply the am gm inequality to get: (abc)^1/3 <= 1 -----------------(1)

Raising both sides to the power 12

hence (abc)^4 <= 1^12=1 -----------------(2)

similarly for a^3 x b^2 + b^3 x c^2 + c^3 x a ^2 :

we get ( a^3 x b^2 + b^3 x c^2 + c^3 x a ^2 )/3 >= (abc)^5/3 = 1.

=> ( a^3 x b^2 + b^3 x c^2 + c^3 x a ^2 )>=3

Consider the case where (a^3 x b^2 + b^3 x c^2 + c^3 x a ^2 ) =3

multiply it with (1)

we get abc(a^3 x b^2 + b^3 x c^2 + c^3 x a ^2 ) <=1 x 3 =3 -------------(3)

adding (2) and (3) we obtain our answer = 1 + 3 = 4