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# RMO practice inequalities #1

Prove that, $\sqrt{2}<\dfrac{\sqrt{2}+\sqrt{3}}{\sqrt{2+\sqrt{3}}}<2$

1 year ago

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$\sqrt{2+\sqrt{3}} = \dfrac{1}{\sqrt{2}}\sqrt{4+2\sqrt{3}} = \dfrac{1}{\sqrt{2}} (\sqrt{3} +1)$

$\implies \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2+\sqrt{3}}} = \sqrt{2} \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{3} +1}$

Now, $$\sqrt{2} + \sqrt{3} > \sqrt{3} +1$$, which implies that $$\dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{3} +1} >1$$.

So, we get that $$\sqrt{2} \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{3} +1} > \sqrt{2}$$.

We shall prove the right side of the inequality,

Since $$\sqrt{6} + \sqrt{2} > \sqrt{3} + \sqrt{2}$$. Which implies that

$\implies \sqrt{2} (\sqrt{3} + 1 ) > \sqrt{3} + \sqrt{2} \implies \sqrt{2} > \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + 1} \implies \sqrt{2} \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{3} +1} < 2$

Hence proved. $$\blacksquare$$ · 1 year ago