RMO Practice problem 11

Algebra Level 4

Find the minimum value of x 2 + y 2 z + y 2 + z 2 x + x 2 + z 2 y \frac{x^{2}+y^{2}}{z}+\frac{y^{2}+z^{2}}{x}+\frac{x^{2}+z^{2}}{y} subject to x + y + z = 1012 x+y+z=1012 and x , y , z > 0 x,y,z>0 .


The answer is 2024.

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Shivam Jadhav by Shivam Jadhav , 22, India

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2 solutions

Duybao Van
Nov 2, 2015

P = x 2 + y 2 z + y 2 + z 2 x + x 2 + z 2 y \Large \frac{x^{2} + y^{2}}{z} + \frac{y^{2} + z^{2}}{x} + \frac{x^{2} + z^{2}}{y}

= x 2 z + y 2 z + y 2 x + z 2 x + x 2 y + z 2 y \Large \frac{x^{2}}{z} + \frac{y^{2}}{z} + \frac{y^{2}}{x} + \frac{z^{2}}{x} + \frac{x^{2}}{y} + \frac{z^{2}}{y}

Using Cauchy Schwartz inequality in Engel form, We have

P ⩾ ( x + x + y + y + z + z ) 2 2 x + 2 y + 2 z \Large \geqslant \frac{(x+x+y+y+z+z)^{2}}{2x+2y+2z}

= ( 2 ( x + y + z ) ) 2 2 ( x + y + z ) \Large \frac { { (2(x+y+z)) }^{ 2 } }{ 2(x+y+z) }

= 4 ( x + y + z ) 2 2 ( x + y + z ) \Large \frac { { 4(x+y+z) }^{ 2 } }{ 2(x+y+z) }

= 2 ( x + y + z ) \normalsize 2(x+y+z)

= 2 × 1012 \normalsize 2 \times 1012

= 2024 \Large \boxed{2024}

12 Helpful 0 Interesting 0 Brilliant 0 Confused

yep same way buddy upvoted

Kaustubh Miglani - 5 years, 6 months ago

What is the meaning of "Engel form"......I did it the same way but used "Titu's Lemma"... Are both same??

Samarth Agarwal - 5 years, 6 months ago

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Yeah Titu's Lemma and the Engel Form are the same thing, same form of the inequality :)

Hrishik Mukherjee - 5 years, 6 months ago
Sayandeep Ghosh
May 10, 2016

Yup same way ....that is done

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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