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Bunuel
If x is positive, what is the least possible value of \(\frac{x}{2} + \frac{2}{x}\)?

(A) 1/2
(B) 1
(C) 2
(D) 3
(E) 4


Are You Up For the Challenge: 700 Level Questions

\(\frac{x}{2} + \frac{2}{x} - 2*\sqrt{x/2}\sqrt{2/x} = (\sqrt{x/2}- \sqrt{2/x})^2 >0\)
\(\frac{x}{2} + \frac{2}{x} > 2\)

IMO C
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nick1816
AM ≥ GM

\((\frac{x}{2} + \frac{2}{x})/2 ≥ \sqrt{\frac{x}{2}*\frac{2}{x}}\)

\((\frac{x}{2} + \frac{2}{x})/2 ≥ 1\)

\((\frac{x}{2} + \frac{2}{x}) ≥ 2\)

C

Bunuel
If x is positive, what is the least possible value of \(\frac{x}{2} + \frac{2}{x}\)?

(A) 1/2
(B) 1
(C) 2
(D) 3
(E) 4


Are You Up For the Challenge: 700 Level Questions

Hi, could you please elaborate on your approach? Is there a definite way to solve such min/max problems??
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Assume x and y are 2 positive numbers

\((x-y)^2 ≥ 0\)

\((x+y)^2 - 4xy ≥ 0\)

\((x+y)^2 ≥ 4xy\)

\(\frac{(x+y)^2}{4} ≥ xy\)

\(\frac{x+y}{2} ≥ \sqrt{xy}\), where x,y>0

arithmetic mean ≥ geometric mean



Since x is positive in the question, 2/x and x/2 >0. You can use the above inequality.



Kritisood
nick1816
AM ≥ GM

\((\frac{x}{2} + \frac{2}{x})/2 ≥ \sqrt{\frac{x}{2}*\frac{2}{x}}\)

\((\frac{x}{2} + \frac{2}{x})/2 ≥ 1\)

\((\frac{x}{2} + \frac{2}{x}) ≥ 2\)

C

Bunuel
If x is positive, what is the least possible value of \(\frac{x}{2} + \frac{2}{x}\)?

(A) 1/2
(B) 1
(C) 2
(D) 3
(E) 4


Are You Up For the Challenge: 700 Level Questions

Hi, could you please elaborate on your approach? Is there a definite way to solve such min/max problems??
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Another approach:
Set x/2 + 2/x = y
(x^2+4)/2x=y
x^2-2xy+4=0
x^2-2xy+y^2-y^2+4=0
(x-y)^2-y^2+4=0
Since (x-y)^2 min = 0
=> y^2=4
=>y=2
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