To Prove the Given Inequality

Given two positive numbers a and b, we define the root mean square as follows:

sqrt{(a^2 + b^2)/2}

Note: The radicand is the fraction
(a^2 + b^2)/2

Note: The arithmetic mean is defined as
(a + b)/2

Prove the following inequality is valid for all positive numbers a and b:

(a + b)/2 is less than or equal to
sqrt{(a^2 + b^2)/2}

(a+b)/2 ≤ √[(a²+b²)/2]

(a+b)²/4 ≤ (a²+b²)/2

(a²+b²+2ab)/4 ≤ (a²+b²)/2

(a²+b²)/4 +ab/2 ≤ (a²+b²)/2

ab/2 ≤ (a²+b²)/4 (subtract (a²+b²)/4 from both sides)

2ab ≤ (a²+b²)

0 ≤ a²+b²-2ab

0 ≤ (a-b)²

hence proved because it is POSITIVE values of a and b

(a + b)/2 ≤ √((a^2 + b^2)/2), a and b positive

square both sides

(a^2 + b^2 + 2ab)/4 ≤ (a^2 + b^2)/2 (multiply by 4 both sides

a^2 + b^2 + 2ab ≤ 2a^2 + 2b^2

a^2 + b^2 + 2ab - 2a^2 - 2b^2 ≤ 0

- a^2 - b^2 + 2ab ≤ 0

- (a^2 + b^2 - 2ab) ≤ 0

- (a - b)^2 ≤ 0

which is true for all a,b

(a-b)^2+ (a+b)^2=2a^2+2b^2 and since (a-b)^2>=0, (a+b)^2<=2a^2+2b^2
so [(a+b)^2]/4 <=(a^2+b^2)/2 . Taking square roots gives the result, remembering
that a and b are positive.

(a + b)/2 <= sqrt((a^2 + b^2)/2)
(a^2 + b^2 + 2ab)/4 <= (a^2 + b^2)/2
a^2 + b^2 + 2ab <= 2a^2 + 2b^2
0 <= a^2 + b^2 - 2ab
0 <= (a - b)^2

proved as any square is greater than or equal to zero

it is impossible to find this result.
because sum of square root is greater