Very easy???

Algebra Level 3

Let a , b , c ≥ 0 a, b, c \geq 0 such that a + b + c = 3 a+b+c = 3 . Find the minimum of a 3 b + 2 + b 3 c + 2 + c 3 a + 2 − a 2 − b 2 − c 2 \frac{a^3}{b+2} + \frac{b^3}{c+2} + \frac{c^3}{a+2} - a^2 - b^2 - c^2


The answer is -2.

Bảo Châu by Bảo Châu , 26, Vietnam

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4 solutions

Shobhit Singh
Nov 22, 2014

Really VERY EASY!!!!!! just 1 min ques..

min. value will be when a=b=c=1. put a=b=c=1 in expression and you will get -2.
1 b'coz a,b,c is positive.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

...Well I guess thats how 75% Of the people got it.....

T sidharth - 4 years, 9 months ago
Rudresh Tomar
Nov 13, 2014

hy guys, my solution. if any flaws do tell me please:

F o r ⟹ a 3 ( c + 2 ) ( a + 2 ) + b 3 ( a + 2 ) ( b + 2 ) + c 3 ( b + 2 ) ( c + 2 ) − ( a 2 + b 2 + c 2 ) ( a + 2 ) ( b + 2 ) ( c + 2 ) ( a + 2 ) ( b + 2 ) ( c + 2 ) t o b e m i n i m u m ⟹ ( a + 2 ) ( b + 2 ) ( c + 2 ) h a s t o b e m a x i m u m n o w , ( a + 2 ) + ( b + 2 ) + ( c + 2 ) 3 ≥ ( a + 2 ) ( b + 2 ) ( c + 2 ) 3 ⟶ [ b y A M ≥ G M ] ⟹ 3 ≥ ( a + 2 ) ( b + 2 ) ( c + 2 ) 3 ⟹ 27 ≥ ( a + 2 ) ( b + 2 ) ( c + 2 ) ∴ t h e m a x v a l u e o f ( a + 2 ) ( b + 2 ) ( c + 2 ) c o m e s 27 w h i c h i s p o s s i b l e o n l y w h e n a = b = c = 1 a f t e r p u t t i n g a = b = c = 1 i n t h e o r i g i n a l e q . t h e m i n i m u m v a l u e c o m e s − 2. For\\ \Longrightarrow \frac { { a }^{ 3 }(c+2)(a+2)+b^{ 3 }(a+2)(b+2)+c^{ 3 }(b+2)(c+2)-({ a }^{ 2 }+b^{ 2 }+c^{ 2 })(a+2)(b+2)(c+2) }{ (a+2)(b+2)(c+2) } \\ to\quad be\quad minimum\\ \Longrightarrow \quad (a+2)(b+2)(c+2)\quad has\quad to\quad be\quad maximum\\ now,\\ \frac { (a+2)+(b+2)+(c+2) }{ 3 } \ge \sqrt [ 3 ]{ (a+2)(b+2)(c+2) } \quad \quad \quad \longrightarrow [by\quad AM\ge GM]\\ \Longrightarrow \quad 3\quad \ge \sqrt [ 3 ]{ (a+2)(b+2)(c+2) } \\ \Longrightarrow \quad 27\quad \ge \quad (a+2)(b+2)(c+2)\\ \therefore \quad the\quad max\quad value\quad of\quad (a+2)(b+2)(c+2)\quad comes\quad 27\quad which\\ \quad \quad \quad is\quad possible\quad only\quad when\quad a=b=c=1\\ after\quad putting\quad a=b=c=1\quad in\quad the\quad original\quad eq.\quad \\ the\quad minimum\quad value\quad comes\quad -2.\quad

0 Helpful 0 Interesting 1 Brilliant 0 Confused
Aditya Tiwari
Nov 13, 2014

Thats it!!!! Did it the same way. Nice solution:)

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Bảo Châu
Aug 21, 2014

Who can write a full solution for this problem?

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Try to use AM-GM and CS repeatedly :)

Krishna Ar - 6 years, 9 months ago

I used the equality case of AM-GM that is I assumed a=b=c=1 can any expert tell me more about this solution

ashutosh mahapatra - 6 years, 8 months ago

I think we can't solve this problem by using AM-GM!!!

Bảo Châu - 6 years, 8 months ago

i also used am gm equality that a=b=c=1 how to solve by method

nikhil jaiswal - 6 years, 8 months ago

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