Sum of fractions, again

Algebra Level 5

x ( y z + 1 ) 2 z 2 ( z x + 1 ) + y ( z x + 1 ) 2 x 2 ( x y + 1 ) + z ( x y + 1 ) 2 y 2 ( y z + 1 ) \large\frac{x(yz+1)^2}{z^2(zx+1)}+\frac{y(zx+1)^2}{x^2(xy+1)}+\frac{z(xy+1)^2}{y^2(yz+1)}

Let x , y , z > 0 x,y,z>0 satisfy x + y + z ≤ 3 2 x+y+z\leq\frac{3}{2} . If the minimum value of the expression above can be written as a b \dfrac{a}{b} which a a and b b are coprime positive integers , calculate the product a b ab .

This problem was taken from Lam Son Highschool grade 10 entrance test for mathematically specialized students


The answer is 30.

P C by P C , 20, Vietnam

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

P C
Jun 10, 2016

Call the expression A, now we rewrite it A = ( y + 1 z ) 2 z + 1 x + ( z + 1 x ) 2 x + 1 y + ( x + 1 y ) 2 y + 1 z A=\frac{\big(y+\frac{1}{z}\big)^2}{z+\frac{1}{x}}+\frac{\big(z+\frac{1}{x}\big)^2}{x+\frac{1}{y}}+\frac{\big(x+\frac{1}{y}\big)^2}{y+\frac{1}{z}} By Titu's Lemma A ≥ ( x + y + z + 1 x + 1 y + 1 z ) 2 x + y + z + 1 x + 1 y + 1 z = x + y + z + 1 x + 1 y + 1 z A\geq\frac{\big(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\big)^2}{x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} We have 1 x + 1 y + 1 z ≥ 9 x + y + z \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq\frac{9}{x+y+z} so x + y + z + 1 x + 1 y + 1 z ≥ x + y + z + 9 x + y + z = x + y + z + 9 4 ( x + y + z ) + 27 4 ( x + y + z ) x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq x+y+z+\frac{9}{x+y+z}=x+y+z+\frac{9}{4(x+y+z)}+\frac{27}{4(x+y+z)} Based on the condition and AM-GM, we have x + y + z + 9 4 ( x + y + z ) ≥ 3 x+y+z+\frac{9}{4(x+y+z)}\geq 3 and 27 4 ( x + y + z ) ≥ 9 2 \frac{27}{4(x+y+z)}\geq\frac{9}{2} ∴ A ≥ 3 + 9 2 = 15 2 = a b \therefore A\geq 3+\frac{9}{2}=\frac{15}{2}=\frac{a}{b} So a b = 30 ab=30 , the equality holds when x = y = z = 1 2 x=y=z=\frac{1}{2}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

how did you convert the equation into that form?

abhishek alva - 5 years ago

Log in to reply

Which equation ?

P C - 5 years ago

Log in to reply

the fist equation

abhishek alva - 5 years ago

Log in to reply

@Abhishek Alva – It's like this x ( y z + 1 ) 2 z 2 ( z x + 1 ) = ( y z + 1 z ) 2 x z + 1 x = ( y + 1 z ) 2 z + 1 x \frac{x(yz+1)^2}{z^2(zx+1)}=\frac{\big(\frac{yz+1}{z}\big)^2}{\frac{xz+1}{x}}=\frac{\big(y+\frac{1}{z}\big)^2}{z+\frac{1}{x}} And same goes for the other 2

P C - 5 years ago

Log in to reply

@P C – is it some method of converting it in that way or just sharp thinking.

abhishek alva - 5 years ago

Log in to reply

@Abhishek Alva – It's just sharp thinking, no method involved

P C - 5 years ago
Son Nguyen
Jun 10, 2016

Do you have a solution? I using U.C.T but it's may have a problem.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

What is U.C.T? And where can I learn it?

Pi Han Goh - 5 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...