Milk It

Algebra Level pending

The following inequality holds itself true for all positive reals x , y , z x,y,z .

x y ( x + y ) + x z ( x + z ) + y z ( y + z ) ≥ C ⋅ x y z xy(x+y) + xz(x+z) + yz(y+z) \geq C \cdot xyz

Evaluate the greatest possible value of the constant C C .

Be sure to include in your answer the procedure used to check the inequality and its equality case. This question is not original.


The answer is 6.

Guilherme Dela Corte by Guilherme Dela Corte , 24, Brazil

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2 solutions

Jubayer Nirjhor
Dec 10, 2014

Direct application of AM-GM: x y ( x + y ) + y z ( y + z ) + z x ( z + x ) = x 2 y + x y 2 + y 2 z + y z 2 + z 2 x + z x 2 ≥ 6 x 2 y ⋅ x y 2 ⋅ y 2 z ⋅ y z 2 ⋅ z 2 x ⋅ z x 2 6 = 6 x 6 y 6 z 6 6 = 6 x y z . \begin{aligned} &~& xy(x+y)+yz(y+z)+zx(z+x) \\ &=& x^2y+xy^2+y^2z+yz^2+z^2x+zx^2 \\ &\ge & 6\sqrt[6]{x^2y\cdot xy^2\cdot y^2z\cdot yz^2\cdot z^2x\cdot zx^2} \\ &=& 6\sqrt[6]{x^6y^6z^6}=6xyz. \end{aligned} Equality occurs when x 2 y = x y 2 = y 2 z = y z 2 = z 2 x = z x 2 x^2y=xy^2=y^2z=yz^2=z^2x=zx^2 that is x = y = z x=y=z . So 6 \boxed{6} is indeed the greatest possible constant.

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x 3 + y 3 + z 3 ≥ 3 1 x + 1 y + 1 z (AM-HM) \dfrac{x}{3} + \dfrac{y}{3} + \dfrac{z}{3} \geq \dfrac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \text{ (AM-HM)}

( x + y + z ) ( 1 x + 1 y + 1 z ) ≥ 9 (x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq 9

x + y + z x + x + y + z y + x + y + z z ≥ 9 \frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{z} \geq 9

y + z x + x + z y + x + y z ≥ 6 \frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z} \geq 6

x y ( x + y ) + x z ( x + z ) + y z ( y + z ) ≥ 6 x y z xy(x+y)+xz(x+z)+yz(y+z) \geq 6xyz

Equality case occurs when x = y = z x=y=z .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

I think you are talking about A.M H.M

U Z - 6 years, 4 months ago

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Edited it: thanks!

Guilherme Dela Corte - 5 years, 7 months ago

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