If and are positive reals, find the minimum value of the expression above.
Submit your answer to 3 decimal places.
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Call the expression E, we've got E + 1 2 = y + z 3 x + 3 + x + z 4 y + 4 + x + y 5 z + 5 ⇔ E + 1 2 = 2 1 ( y + z + x + z + x + y ) ( y + z 3 + x + z 4 + x + y 5 ) Now, applying Cauchy-Schwarz Inequality we get E + 1 2 ≥ 2 ( 3 + 4 + 5 ) 2 ⇔ E ≥ 2 ( 3 + 4 + 5 ) 2 − 1 2 ≈ 5 . 8 0 9