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$
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$