Easy, right?

Algebra Level 5

∑ i = 1 27 ( x i + 1 x i ) 2 \large \sum\limits_{i=1}^{27} \left(x_i+\frac{1}{x_i}\right)^2 If all x x are positive reals satisfying ∑ i = 1 27 x i = 30 \displaystyle \sum\limits_{i=1}^{27} x_i = 30 , find the minimum value of the expression above, round to the nearest integer


The answer is 109.

P C by P C , 20, Vietnam

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1 solution

P C
Feb 4, 2016

Applying Cauchy - Schwarz inequality we have ∑ i = 1 27 ( x i + 1 x i ) 2 ≥ 1 27 ( 30 + ∑ i = 1 27 1 x i ) 2 \sum\limits_{i=1}^{27} (x_i+\frac{1}{x_i})^2\geq\frac{1}{27}(30+\sum\limits_{i=1}^{27} \frac{1}{x_i})^2 Using Titu's Lemma we get ∑ i = 1 27 1 x i ≥ 2 7 2 30 \sum\limits_{i=1}^{27} \frac{1}{x_i}\geq\frac{27^2}{30} So the minimum value is 1 27 ( 30 + 2 7 2 30 ) 2 ≈ 109.2 \frac{1}{27}(30+\frac{27^2}{30})^2\approx 109.2 which is 109 when rounded to the nearest integer

The equality holds when all x = 30 27 x=\frac{30}{27}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

exactly the same way!

Shreyash Rai - 5 years, 4 months ago

I started loving Titu's lemma :). BTW nice solution.

Venkata Karthik Bandaru - 5 years, 4 months ago

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