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Prove that $\sqrt{2}+\sqrt{6}<\sqrt{15}$
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Proof. Assume for a contradiction that $\sqrt{2}+\sqrt{6} \geq \sqrt{15}$
\begin{aligned} &\Longrightarrow(\sqrt{2}+\sqrt{6})^{2} \geq 15 \\ &\Longrightarrow 8+2 \sqrt{12} \geq 15 \\ &\Longrightarrow 2 \sqrt{12} \geq 7 \\ &\Longrightarrow 48 \geq 49 \end{aligned}
The last statement is clearly not true, hence we reached the contradiction. Therefore, we proved that $\sqrt{2}+\sqrt{6}<\sqrt{15}$.
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