Prove for all positive real numbers x and y, if x + y ≤ ( 4 x y ) / ( x + y ), then x = y

Since x + y &gt; 0 , then we obtain ( x + y ) 2 ≤ 4 x y Therefore, x 2 + 2 x y + y 2 ≤ 4 x y which implies that x 2 + y 2 ≤ 2 x y which implies that x 2 − 2 x y + y 2 ≤ 0 and so we get 0 ≤ ( x − y ) 2 ≤ 0 x − y better be 0

Hint: Multiplying by x + y &gt; 0we get ( x + y ) 2 ≤ 4 x y

Since x + y &gt; 0 , then we obtain ( x + y ) 2 ≤ 4 x y Therefore, x 2 + 2 x y + y 2 ≤ 4 x y which implies that x 2 + y 2 ≤ 2 x y which implies that x 2 − 2 x y + y 2 ≤ 0 and so we get 0 ≤ ( x − y ) 2 ≤ 0 x − y better be 0

Hint: Multiplying by x + y &gt; 0we get ( x + y ) 2 ≤ 4 x y

Step by step solution to "Prove for all positive real numbers x and..."

Vincent Norman

Answered question

2022-10-30

Prove for all positive real numbers

x

and

y

, if

x + y \leq (4 x y) / (x + y)

, then

x = y

Answer & Explanation

Annabella Ferguson

Beginner2022-10-31Added 15 answers

Since

x + y > 0

, then we obtain

(x + y)^{2} \leq 4 x y

Therefore,

x^{2} + 2 x y + y^{2} \leq 4 x y

which implies that

x^{2} + y^{2} \leq 2 x y

which implies that

x^{2} - 2 x y + y^{2} \leq 0

and so we get

0 \leq (x - y)^{2} \leq 0

x - y

better be

0

Amina Richards

Beginner2022-11-01Added 2 answers

Hint: Multiplying by

x + y > 0

we get

(x + y)^{2} \leq 4 x y

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