Consider the following problem:Let a,b>0a,b>0 be real numbers such that a+b≥1a+b\geq1. Find the minimum value of the expression A=ab+1abA=ab+{1\over ab}

Now using AM-GM Inequality: A=ab+1ab≥2√ab×1ab=2A=ab+{1\over ab}\geq 2\sqrt{ab\times \frac{1}{ab} }=2 However notice that the equality occurs iff ab=1ab=1 which contradicts the fact that a+b≥1a+b\geq 1.How should I rsolve this contradiction? Is there a way to use AM-GM Inequality to solve such kind of problems?

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How does ab=1a b = 1 contradict a+b≥1a+b \ge 1\;?

– dxiv

2 days ago

Indeed, ab=1ab=1 in fact implies a+b>1a+b>1 (and hence a+b≥1a+b\ge 1) under the given assumptions, since at least one of aa or bb is at least 11, and the other is positive.

– Joey Zou

2 days ago

@dxiv Can you give a counter example?

– tatan

2 days ago

@tatan a=b=1a=b=1 with ab=1,a+b=2≥1a b = 1, a+b=2\ge 1.

– dxiv

2 days ago

1

@tatan x≥2x \ge 2 does not contradict x≥1x \ge 1. In fact, it implies it.

– dxiv

2 days ago

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1 Answer

1

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For a+b=1a+b=1 by AM-GM ab+1ab=ab+16⋅116ab≥1717√ab(116ab)16=1717√11616(ab)15≥ab+\frac{1}{ab}=ab+16\cdot\frac{1}{16ab}\geq17\sqrt[17]{ab\left(\frac{1}{16ab}\right)^{16}}=17\sqrt[17]{\frac{1}{16^{16}(ab)^{15}}}\geq

≥1717√11616(14)15=174\geq17\sqrt[17]{\frac{1}{16^{16}(\frac{1}{4})^{15}}}=\frac{17}{4}

The equality occurs for a=b=12a=b=\frac{1}{2}

Id est, the answer is 174\frac{17}{4}.

If a+b≥1a+b\geq1 so the answer is 22 by your AM-GM.

1

Yeah this is true if a+b≤1a+b\leq 1

– arberavdullahu

2 days ago

That’s creative, but 1⋅1+11⋅1=2<1741 \cdot 1 + \frac {1}{1 \cdot 1} = 2 \lt \frac{17}{4}. P.S. (after the edit to the answer) Where does a+b=1a+b=1 figure in the question? – dxiv 2 days ago