Struggling With Inequality Question (1 Viewer)

Jacobagel · Apr 4, 2019

Prove that a+sqrt(a) is greater than or equal to sqrt[a(a+1)]

HeroWise · Apr 4, 2019

its called the Art of BS

You look at the answer and its certain that $a\geq0$ .

Ok then $2\sqrt{a^3}\geq0$

$a^2+ a + 2\sqrt{a^3} \geq a^2 +a$

$(a+ \sqrt{a})^2 \geq a(a+1)$

$\sqrt{(a+ \sqrt{a})^2} \geq \sqrt{a(a+1)}$

$(a+ \sqrt{a}) \geq \sqrt{a(a+1)}\\ $since its positive, we dont worry of the different signs$$

Jacobagel · Apr 4, 2019

Cheers mate.

Struggling With Inequality Question (1 Viewer)

Jacobagel

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HeroWise

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Jacobagel

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