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Consider a function $$y=(x^3+\frac{1}{x^3})-6(x^2+\frac{1}{x^2})+3(x+\frac{1}{x})$$ defined for real $x>0$. Letting $t=x+\frac{1}{x}$ gives: $$y=t^3-6t^2+12$$

Here it holds that $$t=x+\frac{1}{x}\geq2$$

My question is: how do I know that $t=x+\frac{1}{x}\geq2$ ? I want to know how to get to this point without previouly knowing that $t=x+\frac{1}{x}\geq2$

Renato
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  • Yes, I'm sorry for not noticing it. Do you have any tips about how I can see that the AM GM method can be used? – Renato Aug 04 '20 at 13:53
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    If the product of the terms is 'nicer' than the sum, then it may be worth applying AM-GM. There are a ton of classical inequalities which are useful only in specific situations, so in general, it just comes down to practice. There are also methods using calculus, which I assume you are familiar with based on your other question. – Toby Mak Aug 04 '20 at 13:55

3 Answers3

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Because by AM-GM $$x+\frac{1}{x}\geq2\sqrt{x\cdot\frac{1}{x}}=2.$$ Your calculation of $y$ is right: $$y=t^3-3t-6(t^2-2)+3t=t^3-6t^2+12$$ and you got it without using $t\geq2$.

Michael Rozenberg
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We have that for $x>0$

$$x+\frac{1}{x}\geq2 \iff x\cdot x+x\cdot \frac{1}{x}\geq x\cdot 2 \iff x^2-2x+1\ge 0 \iff (x-1)^2 \ge 0$$

and the equality holds if and only if $x=1$.

user
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hint: use AM GM inequality ,a+b/2 is greater than or equal to root(ab) for positive a,b

Ginger bread
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