Inequalities question (1 Viewer)

apollo1 · Sep 21, 2011

Prove the following for all real numbers x =/= 0

i) $x^{2}+\frac{1}{x^{2}}\geq 2$

ii) $x^{2}+\frac{1}{x^{2}}\geq x+\frac{1}{x}$

ive done the first use AM-GM but i dont no how to do second. halp plz.

AAEldar · Sep 21, 2011

apollo1 said:
Prove the following for all real numbers x =/= 0

i) $x^{2}+\frac{1}{x^{2}}\geq 2$

ii) $x^{2}+\frac{1}{x^{2}}\geq x+\frac{1}{x}$

ive done the first use AM-GM but i dont no how to do second. halp plz.

ii)
$x^{2}+\frac{1}{x^{2}}\geq x+\frac{1}{x} \\\\ x^2+\frac{1}{x^2}-x-\frac{1}{x} \\\\ =x(x-1)-\frac{1}{x^2}(x-1) \\\\ =(x-1)(x-\frac{1}{x^2}) \\\\ \text{if }x<0\text{, both negative hence }>0 \\\\ \text{if }x=1\text{, }(1-1)=0\text{ hence}=0 \\\\ \text{if }x>0\text{, both positive hence }>0 \\\\ \therefore(x-1)(x-\frac{1}{x^2})\geq0 \\\\ \therefore x^2+\frac{1}{x^2}\geq x+\frac{1}{x}$

That's how I'd do it anyway.

Drongoski · Sep 21, 2011

I hesitated posting my solution below as I think there is a far simpler and more elegant way.

$x^2 + \frac {1}{x^2} - x - \frac {1}{x} =\frac {x^4 + 1 -x-x^3}{x^2} = \frac {x^3(x-1)-(x-1)}{x^2} = \frac {(x-1)(x^3-1)}{x^2}$

$= \frac {(x-1)^2(x^2+x+1)}{x^2} = \frac {(x-1)^2([x+\frac {1}{2}]^2) + \frac{3}{4}}{x^2} = \frac {\geq 0 \times \frac {3}{4}}{x^2} \geq 0$

Hence required result.

apollo1 · Sep 21, 2011

AAEldar said:
ii)
$x^{2}+\frac{1}{x^{2}}\geq x+\frac{1}{x} \\\\ x^2+\frac{1}{x^2}-x-\frac{1}{x} \\\\ =x(x-1)-\frac{1}{x^2}(x-1) \\\\ =(x-1)(x-\frac{1}{x^2}) \\\\ \text{if }x<1\text{, both negative hence }>0 \\\\ \text{if }x=1\text{, }(1-1)=0\text{ hence}=0 \\\\ \text{if }x>1\text{, both positive hence }>0 \\\\ \therefore(x-1)(x-\frac{1}{x^2})\geq0 \\\\ \therefore x^2+\frac{1}{x^2}\geq x+\frac{1}{x}$

That's how I'd do it anyway.

how do you deal with the condition that x is not equal to zero?

AAEldar · Sep 21, 2011

apollo1 said:
how do you deal with the condition that x is not equal to zero?

The question stated that it's for x not equal to zero, which is obvious since the x is in the denominator so you don't have to worry about it. There was a slight error though that I just fixed up.

hup · Sep 24, 2011

Inequalities question (1 Viewer)

apollo1

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apollo1

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hup

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