Mathematics: Prove (x^2)+1/(x^2 )≥x + (1/x )≥2 for x 0

Prove (x^2)+1/(x^2 )≥x + (1/x )≥2 for x 0 - id2121203070764, ryansingl19

Ответ:

SamaraP

06.01.2022

Mathematics

First of all, let's prove that

$x+\dfrac{1}{x}\geq 2,\quad x0$

We can rewrite the fraction as

$\dfrac{x^2+1}{x}$

And the inequality as

$\dfrac{x^2+1}{x}-2\geq 0$

Again, we can rewrite as

$\dfrac{x^2-2x+1}{x}\geq 0 \iff \dfrac{(x-1)^2}{x}\geq 0$

The numerator is a square, so it is always greater than or equal to zero.

The denominator is positive, because we're assuming x>0.

So, this is a fraction between positive numbers, and it is always positive, and the first part is finished.

As for the second part, we have

$x^2+\dfrac{1}{x^2} \geq x+\dfrac{1}{x} \iff x^2+\dfrac{1}{x^2}-x-\dfrac{1}{x} \geq 0 \iff \dfrac{(x - 1)^2 (x^2 + x + 1)}{x^2} \geq 0$

The numerator is not the product of a square (always greater than or equal to zero) and a quadratic equation with no solutions and positive leading coefficient (always positive), so it's always greater than or equal to zero. The denominator is x^2, and since x>0 by hypothesis, it is positive. This ends the proof.

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Prove (x^2)+1/(x^2 )≥x + (1/x )≥2 for x> 0

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