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zmudz
 one year ago
Prove that
\(\sqrt{ \frac{2x^2  2x + 1}{2} } \geq \frac{1}{x + \frac{1}{x}}\)
for \(0 < x < 1.\)
zmudz
 one year ago
Prove that \(\sqrt{ \frac{2x^2  2x + 1}{2} } \geq \frac{1}{x + \frac{1}{x}}\) for \(0 < x < 1.\)

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thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0Square both side first?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Fix up the RHS first.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Note that \(x+1/x \geq 2\) (by arithmetic mean > geometric mean or simply by completing the square) Also, \(2x^2  2x + 1 = (1x)^2 + x^2\) Now, use the fact that the rootmeans square is greater than the arithmetic mean, \(\sqrt{\frac{2x^2  2x + 1}{2}} = \sqrt{\frac{(1x)^2 + x^2}{2}} \geq \frac{(1x) + x}{2} = \frac{1}{2}\) Multiply the above two inequalities to get what is required.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@adxpoi I would like to know about the socalled "arithmetic mean> geometric mean". Would you please explain me?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0The problem is easily proving by "square both sides" and do backward of the flow. I would like to learn a new method.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Arithmetic mean is\[\frac{ x_1+x_2+x_3+...+x_n }{ n }\]while geometric mean is\[\sqrt[n]{x_1x_2x_3...x_n}\]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sure. The AM > GM inequality states that for \(n\) nonnegative real numbers \(x_1, x_2,..., x_n\) the following holds: \[\frac{x_1+x_2+...+x_n}{n} \geq (x_1.x_2....x_n)^{\frac{1}{n}}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Thank you. I will read it carefully. @thomas5267

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A possible proof is by induction, first prove it for two numbers, easily done. Then try to extend it by induction.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@adxpoi Yes, I will. Thank you.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Alternatively you may show that the max value of right hand side is \(\frac{1}{2}\) and min value of left hand side is \(\frac{1}{2}\). They occur at different places, so the inequality follows.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0thats exactly same as your first proof, but I was looking at it more from vertex of quadratic equation side.. :)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 I looooooove your max/min method.!! Wow!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Cumbersome, but doesn't require knowledge of the theorems mentioned above.\[\sqrt{\frac{ 2x^22x+1 }{ 2 }} \ge \frac{ 1 }{ x+\frac{ 1 }{ x } }\]\[\sqrt{\frac{ 2\left( x^2x+\frac{ 1 }{ 2 } \right) }{ 2 }} \ge \frac{ x }{ x^2+1 }\]\[x^2x+\frac{ 1 }{ 2 } \ge \frac{ x^2 }{ x^4+2x^2+1 }\]\[x^6x^5+\frac{ 5 }{ 2 }x^42x^3+x^2x+\frac{ 1 }{ 2 } \ge 0\]\[\left( x^6+\frac{ 5 }{ 2 }x^4+x^2+\frac{ 1 }{ 2 } \right)  \left( x^5+2x^3+x \right) \ge 0\]\[x=0 \rightarrow \frac{ 1 }{ 2 } \ge 0\]\[x=1 \rightarrow 0 \ge 0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Excuse me. Last line above \[x=1 \rightarrow 1 \ge 0\]should read

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 isn't the minimum of the LHS \(\sqrt\frac{1}{2}\) ? If so, how does that affect the max/min proposition?
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