Áp dụng BĐT Bunhia thì: \sin x+\cos x \leq \sqrt{2(\sin^2x+\cos^2 x)}=\sqrt{2}Ta có: $P=t+\frac{4}{t^2}+\frac{4}{t} (t=\sin x+\cos x \le\sqrt{2}) =t+\frac{2}{t}+\frac{4}{t^2}+\frac{2}{t}\ge2\sqrt{t.\frac{2}{t}}+\frac{4}{\sqrt{2}^2}+\frac{2}{\sqrt{2}}=4+3\sqrt{2}Dấu bằng có khi x=\pi/4$
Áp dụng BĐT Bunhia thì:
\sin x+\cos x \leq \sqrt{2(\sin^2x+\cos^2 x)}=\sqrt{2}P=\sin x+\cos x+\frac{1+\sin x+\cos x}{\sin x.\cos x}+2\ge t+\frac{1+t}{\frac{t^2}{4}}Ta có: $P
\ge t+\frac{4}{t^2}+\frac{4}{t} (t=\sin x+\cos x \le\sqrt{2})
=t+\frac{2}{t}+\frac{4}{t^2}+\frac{2}{t}\ge2\sqrt{t.\frac{2}{t}}+\frac{4}{\sqrt{2}^2}+\frac{2}{\sqrt{2}}=4+3\sqrt{2}
Dấu bằng có khi x=\pi/4$