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

Note by Adarsh Kumar 2 years, 8 months 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. \(\blacksquare\)

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@Dev Sharma @Harsh Shrivastava @Nihar Mahajan

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TopNewest\[\sqrt{2+\sqrt{3}} = \dfrac{1}{\sqrt{2}}\sqrt{4+2\sqrt{3}} = \dfrac{1}{\sqrt{2}} (\sqrt{3} +1)\]

\[\implies \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2+\sqrt{3}}} = \sqrt{2} \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{3} +1}\]

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. \(\blacksquare\)

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@Dev Sharma @Harsh Shrivastava @Nihar Mahajan

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