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I am trying to prove that $$\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $$

My attempt was to suppose that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} \geq \frac{1}{10}$, and show that it was absurd.

Here's what I did:

Suppose that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} \geq \frac{1}{10} $, then $2 - \sqrt{3} \geq \frac{2\sqrt{3}}{10}$ thus:

$$\frac{20 -10 \sqrt{3}}{10} \geq \frac{2 \sqrt{3}}{10} \implies 20 -\sqrt{300} \geq 2 \sqrt{3} \implies 10 - \sqrt{\frac{300}{4}}\geq \sqrt{3} \implies 10 - \sqrt{75} \geq \sqrt{3} \implies \frac{10}{\sqrt{3}}- \sqrt{25} \geq 1 \implies \frac{10}{\sqrt{3}}- 5 \geq 1$$

Now I elevate everything to the square:

$$\frac{100}{3} - \frac{100}{\sqrt{3}} + 25 \geq 1 \implies \frac{100}{3} - \frac{100 \sqrt{3}}{3} + \frac{75}{3} \geq 1 \implies \frac{100(1-\sqrt{3}) +75}{3} \geq \frac{3}{3}$$

All is left to show is that $$100(1-\sqrt{3}) +75 \geq 3$$

And here, I don't know where to go, and it doesn't seem as I simplified the problem.

Any hints of help will be appreciated

  • 0
    $$\frac{2-\sqrt{3}}{2\sqrt{3}}=\frac{2\sqrt{3}-3}{2\times 3}=\frac{\sqrt{3}}3-\frac 12$$ – 2017-01-29
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    Multiply top and bottom by $2+\sqrt3$; then argue that $4\sqrt3+6>10$. – 2017-01-29

5 Answers 5

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The inequality is iff \begin{aligned} 20-10\sqrt{3}<2\sqrt{3}\iff 20<12\sqrt{3}\iff 400<144\times3=432 \end{aligned} which is clearly true.

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    the last iff because $t \mapsto t^3$ and $t \mapsto t^{\frac{1}{3}}$ are strictly growing on $[0 , \infty)$. – 2017-01-29
3

$$ \frac{2-\sqrt 3}{2 \sqrt3} = \frac{1}{\sqrt 3} - \frac{1}{2} $$ Now, suppose we assume that the opposite of what is given is true. We get a contradiction: $$ \frac{1}{\sqrt 3} - \frac 12 \geq \frac 1{10} \implies\frac{1}{\sqrt 3} \geq \frac{3}{5} \implies \frac{1}{3} \geq \frac{9}{25} \implies 25 \geq 27 $$

which is not true. Hence, we are done.

1

multiplying by $10$ and $$2\sqrt{3}$$ we get $$20-10\sqrt{3}<2\sqrt{3}$$ and we get $$20<12\sqrt{3}$$ dividing by $$4$$ and we have $$5<3\sqrt{3}$$ this is true since we have $$25<27$$

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If you multiply the numerator and denominator by $2+\sqrt3$, you get $\dfrac1{4\sqrt3+6}$. Since $\sqrt3>1$

$$4\sqrt3+6>4(1)+6$$ $$4\sqrt3+6>10$$ $$\frac1{10}>\frac1{4\sqrt3+6}$$

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$\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} \iff 2-\sqrt{3} < \sqrt{3}/5 \iff 2 < 6\sqrt{3}/5 \iff 4 < 108/25 $ which is true.