3 Applications of the Cauchy Schwarz

Algebra Level 3

Non-negative real numbers x x , y y , and z z are such that x 2 + y 2 + z 2 = 3 x^2+y^2+z^2=3 . Calculate the square of the maximum value of the expression below.

x x 2 + y + z + y y 2 + z + x + z z 2 + x + y . \frac{x}{\sqrt{x^2+y+z}}+\frac{y}{\sqrt{y^2+z+x}}+\frac{z}{\sqrt{z^2+x+y}}.

Hint: Look at the title.

Source: Ukrainian Mathematical Olympiad 2008


The answer is 3.

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1 solution

Nitin Kumar
Sep 13, 2020

From the Cauchy- Schwarz inequality, we have ( x 2 + y 2 + z 2 ) 2 = 3 ( x 2 + y 2 + z 2 ) ( x + y + z ) 2 . (x^2+y^2+z^2)^2=3(x^2+y^2+z^2) \geq (x+y+z)^2. This gives, x 2 + y 2 + z 2 x + y + z . \boxed{x^2+y^2+z^2 \geq x+y+z}. Again using the Cauchy- Schwarz inequality, we have ( x 2 + y + z ) ( 1 + y + z ) ( x + y + z ) 2 , (x^2+y+z)(1+y+z) \ge (x+y+z)^2, and their symmetric equivalents. Hence, it suffices to prove that x 1 + y + z + y 1 + z + x + z 1 + x + y x + y + z 3 . \frac{x\sqrt{1+y+z}+y\sqrt{1+z+x}+z\sqrt{1+x+y}}{x+y+z} \leq \sqrt{3}. Now using Cauchy- Schwarz inequality for the third time, we have ( x 1 + y + z + y 1 + z + x + z 1 + x + y ) 2 = ( x x + x y + x z + y y + y x + y z + z z + z x + z y ) 2 ( x + y + z ) ( x + y + z + 2 x y + 2 y z + 2 z x ) ( x + y + z ) ( x 2 + y 2 + z 2 + 2 x y + 2 y z + 2 z x ) = ( x + y + z ) 3 . (x\sqrt{1+y+z}+y\sqrt{1+z+x}+z\sqrt{1+x+y})^2=(\sqrt{x}\sqrt{x+xy+xz}+\sqrt{y}\sqrt{y+yx+yz}+\sqrt{z}\sqrt{z+zx+zy})^2 \leq (x+y+z)(x+y+z+2xy+2yz+2zx) \leq (x+y+z)(x^2+y^2+z^2+2xy+2yz+2zx)=(x+y+z)^3. This gives, x 1 + y + z + y 1 + z + x + z 1 + x + y x + y + z x + y + z x 2 + y 2 + z 2 = 3 . \frac{x\sqrt{1+y+z}+y\sqrt{1+z+x}+z\sqrt{1+x+y}}{x+y+z} \leq \sqrt{x+y+z} \leq \sqrt{x^2+y^2+z^2}=\sqrt{3}. Hence, the square of the maximum value of the wanted expression is 3. 3. Note that the equality occurs at x = y = z = 1 x=y=z=1 .

Just curious, what values of x x , y y , and z z give the maximum value of 3 \sqrt{3} ?

Elijah L - 9 months ago

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x=y=z=1

Symmetric expressions like this one often have their max (or min, depending on context) values when all variables have the same value.

Richard Desper - 8 months, 3 weeks ago

The values would be all of the 3 variables to be unity.

Nitin Kumar - 9 months ago

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So x = y = z = 1 3 4 x = y = z = \frac{1}{\sqrt[4]{3}} ?

Elijah L - 9 months ago

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No, it's x = y = z = 1 x=y=z=1 .

Nitin Kumar - 9 months ago

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@Nitin Kumar Look at the boxed inequality to see why.

Nitin Kumar - 9 months ago

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@Nitin Kumar Then... x 2 + y 2 + z 2 = 3 x^2 + y^2 + z^2 = 3 , not 3 \sqrt{3} ...

Elijah L - 9 months ago

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@Elijah L Ok, yeah,true. Silly me.

Nitin Kumar - 9 months ago

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@Nitin Kumar Is it possible to edit the question so that it's more accurate?

Elijah L - 9 months ago

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@Elijah L Ok, I will. Its been edited.

Nitin Kumar - 9 months ago

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@Nitin Kumar Thank you very much!

Elijah L - 9 months ago

x 2 + y 2 + z 2 = 3 ( x 2 + y 2 + z 2 ) x 2 + y 2 + z 2 = 0 \red{x^2+y^2+z^2 = 3(x^2+y^2+z^2 ) \implies x^2+y^2+z^2 = 0}

Chew-Seong Cheong - 9 months ago

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Oh yeah, there should have been a square in the LHS, i have edited it. Sorry about that!

Nitin Kumar - 9 months ago

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