What positive number added to its reciprocal gives the minimum sum?
x+1x→minCauchys inequality:x+1x⩾2x⋅1x=2The equality holds for x=1x⇒x=1Answer:1x+\frac{1}{x}\rightarrow \min \\Cauchys\,\,inequality:\\x+\frac{1}{x}\geqslant 2\sqrt{x\cdot \frac{1}{x}}=2\\The\,\,equality\,holds\,for\,\,x=\frac{1}{x}\Rightarrow x=1\\Answer: 1x+x1→minCauchysinequality:x+x1⩾2x⋅x1=2Theequalityholdsforx=x1⇒x=1Answer:1
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