Minima and Maxima by Shivam jha

Algebra Level 3

For real numbers r r , s s , and t t such that 1 ≤ r ≤ s ≤ t ≤ 4 1 \le r \le s \le t \le 4 , determine the minimum value of

( r − 1 ) 2 + ( s r − 1 ) 2 + ( t s − 1 ) 2 + ( 4 t − 1 ) 2 \large (r-1)^2+ \left( \frac sr-1\right)^2+ \left(\frac ts-1 \right)^2+ \left(\frac 4t-1\right)^2


The answer is 0.686.

Aditi Jha by Aditi Jha , 17, India

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1 solution

Chew-Seong Cheong
Feb 10, 2018

By Titu's lemma :

( r − 1 ) 2 + ( s r − 1 ) 2 + ( t s − 1 ) 2 + ( 4 t − 1 ) 2 ≥ 1 4 ( r − 1 + s r − 1 + t s − 1 + 4 t − 1 ) 2 \begin{aligned} (r-1)^2 + \left(\frac sr -1\right)^2 + \left(\frac ts -1\right)^2 + \left(\frac 4t -1\right)^2 \ge \frac 14 \left( r-1 + \frac sr -1 + \frac ts -1 + \frac 4t -1\right)^2 \end{aligned}

By AM-GM inequality :

1 4 ( r + s r + t s + 4 t − 4 ) 2 ≥ 1 4 ( 4 4 4 − 4 ) 2 = 4 ( 2 − 1 ) 2 ≈ 0.686 \begin{aligned} \frac 14 \left({\color{#3D99F6} r + \frac sr + \frac ts + \frac 4t} -4\right)^2 \ge \frac 14 \left({\color{#3D99F6} 4\sqrt[4]4} - 4\right)^2 = 4(\sqrt 2-1)^2 \approx \boxed{0.686} \end{aligned}

Equality occurs when r = 2 r=\sqrt 2 , s = 2 s = 2 and t = 2 2 t = 2\sqrt 2 .

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