If x, y, z0 and x+y+z=1, then find the least value of E=(2 x/1-x)+(2 y/1-y)+(2 z/1-z).

Question

If x, y, z>0 and x+y+z=1, then find the least value of E=(2 x/1-x)+(2 y/1-y)+(2 z/1-z). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

E=2[(x/1-x)+(y/1-y)+(z/1-z)]=2[(1/1-x)+(1/1-y)+(1/1-z)-3] Now, ((1-x)+(1-y)+(1-z)/3) ≥ (3/(1)1-x+(1)1-y+(1/1-z) (2/3) ≥ (3/(1)1-x+(1)1-y+(1/1-z) ⇒ (1/1-x)+(1/1-y)+(1/1-z) ≥ (9/2) ⇒ E ≥ 2[(9/2)-3]=3 ⇒ E / min )=3

Q. If x,y,z>0 and x+y+z=1, then find the least value of E=1−x2x​+1−y2y​+1−z2z​.

E=2[1−xx​+1−yy​+1−zz​]=2[1−x1​+1−y1​+1−z1​−3] Now, 3(1−x)+(1−y)+(1−z)​≥1−x1​+1−y1​+1−z1​3​ 32​≥1−x1​+1−y1​+1−z1​3​ ⇒1−x1​+1−y1​+1−z1​≥29​ ⇒E≥2[29​−3]=3⇒Emin​=3

Q. If $x, y, z > 0$ and $x + y + z = 1$ , then find the least value of $E = \frac{2 x}{1 - x} + \frac{2 y}{1 - y} + \frac{2 z}{1 - z}$ .