Blogs của Phúc

Cho $x,y,z> 0$  thỏa mãn $x+y+z=1$. Chứng minh rằng: $$P=\frac{xy}{\sqrt{zx+yz}}+\frac{yz}{\sqrt{xy+zx}}+\frac{zx}{\sqrt{yz+xy}}\leqslant \frac{\sqrt{2}}{2}$$                                                                           Giải Ta có: $P\leqslant \frac{\sqrt2}{2}$ $$\Leftrightarrow xy\sqrt{(xy+xz)(xy+yz)}+yz\sqrt{(yz+xy)(yz+zx)}+zx\sqrt{(zx+xy)(zx+yz)}$$ $$\leq \frac{\sqrt2}{2}\sqrt{(xy+yz)(yz+zx)(zx+xy)}$$ $$\Leftrightarrow xy.\sqrt{x^{2}y^{2}+xyz}+yz.\sqrt{y^{2}z^{2}+xyz}+zx.\sqrt{z^{2}x^{2}+xyz}$$ $$\leqslant \frac{\sqrt2}{2}.\sqrt{xyz}.\sqrt{xy+yz+zx-xyz}$$ $$P^{2}=\frac{x^2y^2}{zx+zy}+\frac{y^2z^2}{xy+xz}+\frac{z^2x^2}{yx+yz}+2(\frac{x^2}{\sqrt{x+yz}}+\frac{y^2}{\sqrt{y+xz}}+\frac{z^2}{\sqrt{z+xy}})$$

b) Ta có: $$sinx-cosx=-2sin\frac{\pi}{4}sin(\frac{\pi}{4}-x)$$ Nên ĐKXĐ: $$-2sin\frac{\pi}{4}sin(\frac{\pi}{4}-x) > 0$$ $$\Leftrightarrow sin(\frac{\pi}{4}-x)<0$$ $$\Leftrightarrow -\pi+k2\pi<\frac{\pi}{4}-x<k2\pi$$ $$\Leftrightarrow \frac{5\pi}{4}+k2\pi>x>\frac{\pi}{4}+k2\pi$$ c) ĐKXĐ: $$cosx\leqslant 0$$ $$\Leftrightarrow \frac{\pi}{2}+k2\pi\leq x\leq \frac{3\pi}{2}+k2\pi$$ d) $$sin(cosx)\geq 0$$ $$\Leftrightarrow k2\pi\leq cosx\leq \pi+k2\pi$$ Mặt khác: $$-1\leqslant cosx\leqslant 1 $$