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How would you prove the following:

(a) [tex] x + \frac{1}{x} \geq 2, x > 0 [/tex]

(b) [tex] x + \frac{1}{x} \leq -2, x < 0 [/tex]

(c) [tex] |x+\frac{1}{x}| \geq 2, x\neq 0. [/tex]

For all of these inequalities would I simply solve for x, or would I have to use things like the triangle inequality of Schwarz's Inequality?

Ok I was also given some harder problems with the same concepts:Prove the following inequalities:

(i) [tex] x^2 + xy + y^2\geq 0 [/tex]

(ii) [tex] x^{2n} +x^{2n-1}y + x^{2n-2}y^2+ ... + y^{2n} \geq 0 [/tex]

(iii) [tex] x^4 - 3x^3 + 4x^2 - 3x + 1 \geq 0. [/tex]

Ok for (i) we can factor [tex]\frac{x^3-y^3}{x-y} [/tex]. If [tex] x > y, x < y [/tex] this expression is positive. For [tex] x = y [/tex] we have [tex] 3x^2 [/tex] which is positive.

For (ii) we have the same thing except [tex] \frac{x^{2n+1} - y^{2n+1}}{x-y} [/tex].

For (iii) [tex] x^4 - 3x^3 + 4x^2 - 3x + 1 \geq 0

\ [/tex][tex] x^4 - 3x^3 + 4x^2 - 3x + 1 = (x-1)^{2}(x^2-x+1) [/tex]

[tex]x^2 - x + 1 = (x-\frac{1}{2})^2 + \frac{3}{4} \geq \frac{3}{4} [/tex]

Would the above problems be similar to these?

Thanks

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# Homework Help: Proving Inequalities

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