Making A Big Product Small

Algebra Level 3

( 1 + 1 x ) ( 1 + 1 y ) ( 1 + 1 z ) \left( 1+ \dfrac1x\right) \left( 1 + \dfrac1y \right) \left( 1 + \dfrac 1z \right )

If x , y , x,y, and z z are positive reals with x + y + z = 1 , x + y + z = 1, then what is the minimum value of the expression above?


The answer is 64.

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P C by P C , 20, Vietnam

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3 solutions

P C
Jan 8, 2016

Multiply the expression and we get A = 1 + 1 x y + 1 y z + 1 x z + 1 x + 1 y + 1 z + 1 x y z A=1+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xyz} Using AM-GM we get 1 x y + 1 y z + 1 x z ≥ 4 [ 1 ( x + y ) 2 + 1 ( y + z ) 2 + 1 ( x + z ) 2 ] ≥ 12 [ ( x + y ) ( y + z ) ( x + z ) ] 2 3 ≥ 12 4 ( x + y + z ) 2 9 = 27 \frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\geq4[\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(x+z)^2}]\geq\frac{12}{\sqrt[3]{[(x+y)(y+z)(x+z)]^2}}\geq\frac{12}{\frac{4(x+y+z)^2}{9}}=27 1 x y z ≥ 27 ( x + y + z ) 3 = 27 \frac{1}{xyz}\geq\frac{27}{(x+y+z)^3}=27 Using AM-HM we get x + y + z 3 ≥ 3 1 x + 1 y + 1 z ⇔ 1 x + 1 y + 1 z ≥ 9 \frac{x+y+z}{3}\geq\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq9 Finally ⇒ A ≥ 1 + 9 + 27 + 27 = 64 \Rightarrow A\geq1+9+27+27=64 The equality holds when x = y = z = 1 3 x=y=z=\frac{1}{3}

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Alternatively, instead of using a relatively complicated AM-GM to find 1 x y + 1 y z + 1 z x ≥ 27 \frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} \geq 27 , one can find that 1 x y + 1 y z + 1 z x = z x y z + x x y z + y x y z = x + y + z x y z = 1 x y z \frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} = \frac{z}{xyz} + \frac{x}{xyz} + \frac{y}{xyz} = \frac{x + y + z}{xyz} = \frac{1}{xyz} and use a simpler AM-GM to find 1 x y z ≥ 27 ( x + y + z ) 3 = 27 \frac{1}{xyz} \geq \frac{27}{(x+y+z)^3} = 27 as stated in the solution above.

Vincent Tandya - 5 years, 5 months ago
Parth Bhardwaj
Jan 13, 2016

Instead of calculating for so long, cant we just say that the maxima or minima of these problems are when all the variables are equal (if they vary linearly with respect to each other and are followed by same coefficients - in simple words - symmetric equation ), this is because all have the equal importance the equation, thus their values must be equal ???? And then we can check it out for the fact that it is a maxima or a minima ????

Thus x=y=z=1/3 which gives us the answer to be 4 * 4 * 4=64 ....

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Moderator note:

Nope, not at all. This is a very common misconception made.

As an explicit counter example, what is the maximium and minimum of ( x − 1 ) 2 ( y − 1 ) 2 (x-1)^2(y-1)^2 subject to x + y = 3 , x ≥ 0 , y ≥ 0 x + y = 3, x \geq 0, y \geq 0 ? Neither of the answers are 1 16 \frac{1}{16} .

Applying AM-HM inequality first, we obtain

x + y + z 3 = 1 3 ≥ 3 1 x + 1 y + 1 z \frac{x + y + z}{3} = \frac{1}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} ⇔ 1 x + 1 y + 1 z ≥ 9 \Leftrightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq 9 .

Next, applying the given expression using AM-GM inequality yields

1 + 1 x + 1 + 1 y + 1 + 1 z = 12 ≤ 3 ( 1 + 1 x ) ( 1 + 1 y ) ( 1 + 1 z ) 3 1 + \frac{1}{x} + 1 + \frac{1}{y} + 1 + \frac{1}{z} = 12 \leq 3\sqrt[3]{(1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z})} ⇔ ( 1 + 1 x ) ( 1 + 1 y ) ( 1 + 1 z ) ≥ 64 \Leftrightarrow (1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z}) \geq 64 .

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