RMO practice inequalities #1

Prove that, 2<2+32+3<2\sqrt{2}<\dfrac{\sqrt{2}+\sqrt{3}}{\sqrt{2+\sqrt{3}}}<2

#RMO #Inequality

Note by Adarsh Kumar
5 years, 7 months ago

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2+3=124+23=12(3+1)\sqrt{2+\sqrt{3}} = \dfrac{1}{\sqrt{2}}\sqrt{4+2\sqrt{3}} = \dfrac{1}{\sqrt{2}} (\sqrt{3} +1)

    2+32+3=22+33+1\implies \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2+\sqrt{3}}} = \sqrt{2} \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{3} +1}

Now, 2+3>3+1\sqrt{2} + \sqrt{3} > \sqrt{3} +1, which implies that 2+33+1>1\dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{3} +1} >1.

So, we get that 22+33+1>2\sqrt{2} \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{3} +1} > \sqrt{2}.

We shall prove the right side of the inequality,

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

    2(3+1)>3+2    2>3+23+1    22+33+1<2\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

Surya Prakash - 5 years, 7 months ago
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