A problem I posted a long time ago... but no one cares.

Level 2

a; b; c \text{a; b; c} are three positive reals such that ( a + b + c ) ( 1 a + 1 b + 1 c ) = n (a + b + c)\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right) = n and the maximum value of ( a 2 + b 2 + c 2 ) ( 1 a 2 + 1 b 2 + 1 c 2 ) (a^2 + b^2 + c^2)\left(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}\right) is 16 n 16n .

What's the minimum value of ( a 2 + b 2 + c 2 ) ( 1 a 2 + 1 b 2 + 1 c 2 ) (a^2 + b^2 + c^2)\left(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}\right) ?


The answer is 481.5.

Thành Đạt Lê by Thành Đạt Lê , 17, Singapore

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1 solution

Thành Đạt Lê
Apr 14, 2018

We have that ( a + b + c ) ( 1 a + 1 b + 1 c ) (a + b + c)\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right)

= 3 + a b + a c + b c + b a + c a + c b = 3 + \dfrac{a}{b} + \dfrac{a}{c} + \dfrac{b}{c} + \dfrac{b}{a} + \dfrac{c}{a} + \dfrac{c}{b}

= 3 + c a b c + a b b c + a b c a + b c c a + b c a b + c a a b = 3 + \dfrac{ca}{bc} + \dfrac{ab}{bc} + \dfrac{ab}{ca} + \dfrac{bc}{ca} + \dfrac{bc}{ab} + \dfrac{ca}{ab}

= ( a b + b c + c a ) ( 1 a b + 1 b c + 1 c a ) = (ab + bc + ca)\left(\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca}\right) .

16 n + 2 n = ( a 2 + b 2 + c 2 ) ( 1 a 2 + 1 b 2 + 1 c 2 ) + 2 ( a b + b c + c a ) 2 ( 1 a b + 1 b c + 1 c a ) \sqrt{16n} + 2\sqrt{n} = \sqrt{(a^2 + b^2 + c^2)\left(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}\right)} + \sqrt{2(ab + bc + ca)2\left(\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca}\right)}

⟺ 6 n ≤ [ a 2 + b 2 + c 2 + 2 ( a b + b c + c a ) ] [ 1 a 2 + 1 b 2 + 1 c 2 + 2 ( 1 a b + 1 b c + 1 c a ) ] \iff 6\sqrt{n} \le \sqrt{[a^2 + b^2 + c^2 + 2(ab + bc + ca)]\left[\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} + 2\left(\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca}\right)\right]}

⟺ 6 n ≤ ( a + b + c ) ( 1 a + 1 b + 1 c ) = n \iff 6\sqrt{n} \le (a + b + c)\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right) = n

⟺ 6 ≤ n \iff 6 \le \sqrt{n}

⟺ 36 ≤ n \iff 36 \le n . So the equality happens when n = 36 n = 36 .

(The maximum value of the expression ( a 2 + b 2 + c 2 ) ( 1 a 2 + 1 b 2 + 1 c 2 ) (a^2 + b^2 + c^2)\left(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}\right) for any n n given is n ( n − 2 ) 2 n (n - \sqrt 2)^2 , just to let you know.)

Let x = a b + b c + c a x = \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} and y = a c + b a + c b y = \dfrac{a}{c} + \dfrac{b}{a} + \dfrac{c}{b}

We have that ( a + b + c ) ( 1 a + 1 b + 1 c ) = n (a + b + c)\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right) = n

⟺ x + y = n − 3 \iff x + y = n - 3

( a 2 + b 2 + c 2 ) ( 1 a 2 + 1 b 2 + 1 c 2 ) (a^2 + b^2 + c^2)\left(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}\right)

= 3 + a 2 b 2 + b 2 c 2 + c 2 a 2 + a 2 c 2 + b 2 a 2 + c 2 b 2 = 3 + \dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} + \dfrac{c^2}{a^2} + \dfrac{a^2}{c^2} + \dfrac{b^2}{a^2} + \dfrac{c^2}{b^2}

= 3 + ( a b + b c + c a ) 2 − 2 ( a c + b a + c b ) + ( a c + b a + c b ) 2 − 2 ( a b + b c + c a ) = 3 + \left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}\right)^2 - 2\left(\dfrac{a}{c} + \dfrac{b}{a} + \dfrac{c}{b}\right) + \left(\dfrac{a}{c} + \dfrac{b}{a} + \dfrac{c}{b}\right)^2 - 2\left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}\right)

= 3 + x 2 − 2 y + y 2 − 2 x = 3 + x^2 - 2y + y^2 - 2x

= 1 + ( x 2 2 + y 2 2 − x y ) + ( x 2 2 + y 2 2 + x y − 2 x − 2 y + 2 ) = 1 + \left(\dfrac{x^2}{2} + \dfrac{y^2}{2} - xy\right) + \left(\dfrac{x^2}{2} + \dfrac{y^2}{2} + xy - 2x - 2y + 2\right)

= 1 + ( x − y ) 2 2 + ( x + y − 2 ) 2 2 = 1 + \dfrac{(x - y)^2}{2} + \dfrac{(x + y - 2)^2}{2}

≥ 1 + ( x + y − 2 ) 2 2 \ge 1 + \dfrac{(x + y - 2)^2}{2}

= 1 + ( n − 5 ) 2 2 = 1 + \dfrac{(n - 5)^2}{2}

Plugging in n = 36 n = 36 , we have that ( a 2 + b 2 + c 2 ) ( 1 a 2 + 1 b 2 + 1 c 2 ) ≥ 1 + 3 1 2 2 = 481.5 (a^2 + b^2 + c^2)\left(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}\right) \ge 1 + \dfrac{31^2}{2} = 481.5 .

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