Tricky 2

Algebra Level 4

Let a , b , c a,b,c be positive real numbers such that a + b + c = 3 a+b+c=3 . If the minimal value of a b c + 12 a b + b c + c a abc+\dfrac{12}{ab+bc+ca} is A A , what is A 5 \dfrac{A}{5} ?

3 4 2 1 5

Reynan Henry by Reynan Henry , 22, Egypt

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1 solution

Reynan Henry
Jan 14, 2017

We use a well-known inequality a b c ≥ ( a + b − c ) ( b + c − a ) ( c + a − b ) abc\ge (a+b-c)(b+c-a)(c+a-b)

a b c ≥ ( a + b − c ) ( b + c − a ) ( c + a − b ) = ( 3 − 2 a ) ( 3 − 2 b ) ( 3 − 2 c ) = 27 − 18 ( a + b + c ) + 12 ( a b + b c + c a ) − 8 a b c 9 a b c ≥ 27 − 54 + 12 ( a b + b c + c a ) a b c ≥ 4 ( a b + b c + c a ) 3 − 3 \begin{aligned} abc&\ge (a+b-c)(b+c-a)(c+a-b)\\ &=(3-2a)(3-2b)(3-2c)\\ &=27 -18(a+b+c)+12(ab+bc+ca)-8abc \\ 9abc&\ge 27-54+12(ab+bc+ca) \\ abc&\ge \frac{4(ab+bc+ca)}{3}-3\end{aligned}

so

a b c + 12 a b + b c + c a ≥ 4 ( a b + b c + c a ) 3 − 3 + 12 a b + b c + c a ≥ 8 − 3 = 5 \begin{aligned}abc+\frac{12}{ab+bc+ca}&\ge \frac{4(ab+bc+ca)}{3}-3 + \frac{12}{ab+bc+ca}\\ &\ge 8 -3\\ &= 5\end{aligned} by AM-GM. Hence the asnwer is 1 1

6 Helpful 0 Interesting 0 Brilliant 0 Confused

You need to add the fact that all these inequalities are in fact equalities when a = b = c = 1 a=b=c=1 , so that 5 5 is not just a lower bound for your expression, but in fact the minimum value.

Mark Hennings - 4 years, 5 months ago

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