设 $\alpha,\beta\in(0,\dfrac{\pi}{2})$，证明：$\cos\alpha+\cos\beta+\sqrt{2}\sin\alpha\sin\beta\leqslant \dfrac{3\sqrt{2}}{2}$．
【难度】
【出处】
2018年全国高中数学联赛河北省预赛（高二）
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【答案】
略
【解析】
因为 $\begin{matrix}
& \cos \alpha +\cos \beta +\sqrt{2}\sin \alpha \sin \beta\text{=}2\cos \dfrac{\alpha +\beta }{2}\cos \dfrac{\alpha -\beta}{2}+\dfrac{\sqrt{2}}{2}\left[ \cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right) \right] \\
& \leqslant 2\cos \dfrac{\alpha +\beta}{2}+\dfrac{\sqrt{2}}{2}\left[ 1-\cos \left( \alpha +\beta \right) \right] \\
& \text{=}2\cos \dfrac{\alpha +\beta}{2}+\dfrac{\sqrt{2}}{2}\left( 2-2{{\cos }^{2}}\dfrac{\alpha +\beta }{2} \right)\\
& \text{=}\sqrt{2}-\sqrt{2}{{\cos}^{2}}\dfrac{\alpha +\beta }{2}+2\cos \dfrac{\alpha +\beta }{2} \\
&\text{=}\dfrac{3\sqrt{2}}{2}-\sqrt{2}{{\left( \cos \dfrac{\alpha +\beta}{2}-\dfrac{\sqrt{2}}{2} \right)}^{2}}\leqslant \dfrac{3\sqrt{2}}{2} \\
\end{matrix}$，当且仅当 $\begin{cases}\cos\dfrac{\alpha-\beta}{2}=1\\
\cos\dfrac{\alpha+\beta}{2}-\dfrac{\sqrt{2}}{2}=0\\
\end{cases}$ 即 $\alpha=\beta=\dfrac{\pi}{4}$ 时等号成立．故原不等式得证．
& \cos \alpha +\cos \beta +\sqrt{2}\sin \alpha \sin \beta\text{=}2\cos \dfrac{\alpha +\beta }{2}\cos \dfrac{\alpha -\beta}{2}+\dfrac{\sqrt{2}}{2}\left[ \cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right) \right] \\
& \leqslant 2\cos \dfrac{\alpha +\beta}{2}+\dfrac{\sqrt{2}}{2}\left[ 1-\cos \left( \alpha +\beta \right) \right] \\
& \text{=}2\cos \dfrac{\alpha +\beta}{2}+\dfrac{\sqrt{2}}{2}\left( 2-2{{\cos }^{2}}\dfrac{\alpha +\beta }{2} \right)\\
& \text{=}\sqrt{2}-\sqrt{2}{{\cos}^{2}}\dfrac{\alpha +\beta }{2}+2\cos \dfrac{\alpha +\beta }{2} \\
&\text{=}\dfrac{3\sqrt{2}}{2}-\sqrt{2}{{\left( \cos \dfrac{\alpha +\beta}{2}-\dfrac{\sqrt{2}}{2} \right)}^{2}}\leqslant \dfrac{3\sqrt{2}}{2} \\
\end{matrix}$，当且仅当 $\begin{cases}\cos\dfrac{\alpha-\beta}{2}=1\\
\cos\dfrac{\alpha+\beta}{2}-\dfrac{\sqrt{2}}{2}=0\\
\end{cases}$ 即 $\alpha=\beta=\dfrac{\pi}{4}$ 时等号成立．故原不等式得证．
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