Question

Using Identities prove that (x+y+z)^2 ≥ 3(xy+yz+zx) where x,y,z are positive real numbers.​

Answer 1

It is proved that (x + y + z)² ≥ 3(xy + yz + zx) where x, y, and z are positive real numbers.​

Given:
x, y and z are positive real numbers.​

To Prove:

Prove that (x + y + z)² ≥ 3(xy + yz + zx) where x, y, and z are positive real numbers.​

Solution:

We are going to use the AM-GM concept here.

It is given by

Arithmetic Mean ≥ Geometric Mean

AM ≥ GM

If x and y are positive real numbers, then

(x + y)/2 ≥ √(xy)

Squaring both sides, we have

(x + y)² ≥ 4xy

x² + y² + 2xy ≥ 4xy

x² + y² ≥ 2xy ----------------(1)

If y and z are positive real numbers, then

(y + z)/2 ≥ √(yz)

Squaring both sides, we have

(y + z)² ≥ 4yz

y² + z² + 2yz ≥ 4yz

y² + z² ≥ 2yz ----------------(2)

If x and z are positive real numbers, then

(x + z)/2 ≥ √(xz)

Squaring both sides, we have

(x + z)² ≥ 4xz

x² + z² + 2xyz ≥ 4xz

x² + z² ≥ 2xz ----------------(3)

By adding equations (1), (2) and (3) we get

x² + y² + y² + z² + x² + z² ≥ 2xy + 2yz + 2zx

2(x² + y² + z²) ≥ 2(xy + yz + zx)

x² + y² + z² ≥ xy + yz + zx

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx ---------------(4)

We found that x² + y² + z² ≥ xy + yz + zx

Putting the value in equation (4), we get

(x + y + z)² ≥ xy + yz + zx + 2xy + 2yz + 2zx

(x + y + z)² ≥ 3(xy + yz + zx)

It is proved that (x + y + z)² ≥ 3(xy + yz + zx) where x, y, and z are positive real numbers.​

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