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Question: Let x,y,z be positive real numbers, then the least value of $\frac{x(1+y)+y(1+z)+z(1+x)}{\sqrt{xyz}}...

Let x,y,z be positive real numbers, then the least value of x(1+y)+y(1+z)+z(1+x)xyz\frac{x(1+y)+y(1+z)+z(1+x)}{\sqrt{xyz}} is equal to-

A

92\frac{9}{\sqrt{2}}

B

6

C

16\frac{1}{\sqrt{6}}

D

3

Answer

6

Explanation

Solution

The given expression is E=x(1+y)+y(1+z)+z(1+x)xyzE = \frac{x(1+y)+y(1+z)+z(1+x)}{\sqrt{xyz}}.

First, expand the numerator: E=x+xy+y+yz+z+zxxyzE = \frac{x+xy+y+yz+z+zx}{\sqrt{xyz}}

Rearrange the terms in the numerator: E=(x+y+z)+(xy+yz+zx)xyzE = \frac{(x+y+z) + (xy+yz+zx)}{\sqrt{xyz}}

Now, we apply the AM-GM inequality to the terms in the numerator. For positive real numbers x,y,zx, y, z:

  1. x+y+z≥3xyz3x+y+z \ge 3\sqrt[3]{xyz}

Equality holds when x=y=zx=y=z.

  1. xy+yz+zx≥3(xy)(yz)(zx)3xy+yz+zx \ge 3\sqrt[3]{(xy)(yz)(zx)}

xy+yz+zx≥3x2y2z23xy+yz+zx \ge 3\sqrt[3]{x^2y^2z^2}

xy+yz+zx≥3(xyz)2/3xy+yz+zx \ge 3(xyz)^{2/3}

Equality holds when xy=yz=zxxy=yz=zx. Since x,y,zx,y,z are positive, this implies x=y=zx=y=z.

Substitute these inequalities into the expression for EE: E≥3(xyz)1/3+3(xyz)2/3(xyz)1/2E \ge \frac{3(xyz)^{1/3} + 3(xyz)^{2/3}}{(xyz)^{1/2}}

Now, simplify the powers of xyzxyz: E≥3(xyz)1/3(xyz)1/2+3(xyz)2/3(xyz)1/2E \ge 3 \frac{(xyz)^{1/3}}{(xyz)^{1/2}} + 3 \frac{(xyz)^{2/3}}{(xyz)^{1/2}}

E≥3(xyz)1/3−1/2+3(xyz)2/3−1/2E \ge 3 (xyz)^{1/3 - 1/2} + 3 (xyz)^{2/3 - 1/2}

E≥3(xyz)2/6−3/6+3(xyz)4/6−3/6E \ge 3 (xyz)^{2/6 - 3/6} + 3 (xyz)^{4/6 - 3/6}

E≥3(xyz)−1/6+3(xyz)1/6E \ge 3 (xyz)^{-1/6} + 3 (xyz)^{1/6}

Let t=(xyz)1/6t = (xyz)^{1/6}. Since x,y,zx,y,z are positive real numbers, tt must be a positive real number. The inequality becomes:

E≥3t−1+3tE \ge 3t^{-1} + 3t

E≥3(1t+t)E \ge 3 \left( \frac{1}{t} + t \right)

Finally, apply the AM-GM inequality to tt and 1t\frac{1}{t}: For any positive real number tt, t+1t≥2t⋅1tt + \frac{1}{t} \ge 2\sqrt{t \cdot \frac{1}{t}}

t+1t≥21t + \frac{1}{t} \ge 2\sqrt{1}

t+1t≥2t + \frac{1}{t} \ge 2

Equality holds when t=1tt = \frac{1}{t}, which implies t2=1t^2=1. Since t>0t>0, we must have t=1t=1.

Substitute this back into the expression for EE: E≥3(2)E \ge 3(2)

E≥6E \ge 6

To confirm that 6 is the least value, we need to show that equality can be achieved. All the AM-GM steps yield equality if:

  1. x=y=zx=y=z
  2. t=1t=1, which means (xyz)1/6=1(xyz)^{1/6}=1, so xyz=1xyz=1.

If x=y=zx=y=z and xyz=1xyz=1, then x⋅x⋅x=1x \cdot x \cdot x = 1, so x3=1x^3=1. Since xx is a positive real number, x=1x=1. Therefore, when x=y=z=1x=y=z=1, all the conditions for equality are met.

Let's calculate the value of EE when x=y=z=1x=y=z=1: E=1(1+1)+1(1+1)+1(1+1)1⋅1⋅1E = \frac{1(1+1)+1(1+1)+1(1+1)}{\sqrt{1 \cdot 1 \cdot 1}}

E=1(2)+1(2)+1(2)1E = \frac{1(2)+1(2)+1(2)}{1}

E=2+2+21E = \frac{2+2+2}{1}

E=6E = 6

Since the value 6 can be achieved, and we have shown E≥6E \ge 6, the least value of the expression is 6.