If x,y,z are positive numbers such that x+y+z=1, prove xy+yz+zx-\dfrac{9}{4}xyz\leq \dfrac{1}{4}

Collected in the board: Inequality

Steven Zheng posted 2 days ago


Rewrite the inequality to the following form.

(1-\dfrac{9}{4}y)zx\leq \dfrac{1}{4}-y(x+z)

Since x+z=1-y

the inequality is equivalent to

(1-\dfrac{9}{4}y)zx\leq y^2-y+\dfrac{1}{4}=(y-\dfrac{1}{2} )^2

Since x,y,z are positive, 0\leq y\leq 1

If y\geq \dfrac{4}{9}, then the LHS is negative while the RHS is perfect square (y-\dfrac{1}{2})^2. So the inequality is true.

If the inequality becomes equal, then

(1-\dfrac{9}{4}y)zx=(y-\dfrac{1}{2} )^2=0

x = 0, y=\dfrac{1}{2}, z = \dfrac{1}{2} , or z=0, y=\dfrac{1}{2}, x = \dfrac{1}{2}

If y<\dfrac{4}{9}, then 1-\dfrac{9}{4}y>0

Apply arithmetic means inequality

zx\leq \dfrac{(z+x)^2}{4} =\dfrac{(1-y)^2}{4}

Then, what we need to do is to show the following inequality is true

(1-\dfrac{9}{4}y)\dfrac{(1-y)^2}{4}\leq y^2-y+\dfrac{1}{4}

(4-9y)(y^2-2y+1)\leq 16y^2-16y+4

Expand the brackets on the left hand side

4y^2-8y+4-9y^3+18y^2-9y\leq 16y^2-16y+4

Simply and its found that the vertex of the quadratic is positive and the inequality is true

-9y^3+6y^2-y\leq 0

-y(3y-1)^2\leq 0

If the inequality is equal, then x=z and y(3y-1)^2

the we get two solutions,

y = 0, x=z=\dfrac{1}{2}

y=\dfrac{1}{3}, x=z=\dfrac{1}{3}

Steven Zheng posted 2 days ago

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