Bất đẳng thức số 1
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: $P\leqslant \frac{\sqrt2}{2}$ $$\Leftrightarrow xy\sqrt{(xy+xz)(xy+yz)}+yz\sqrt{(yz+xy)(yz+zx)}+zx\sqrt{(zx+xy)(zx+yz)}$$
$$\leq \frac{\sqrt2}{2}\sqrt{(xy+yz)(yz+zx)(zx+xy)}$$
$$\Leftrightarrow xy.\sqrt{x^{2}y^{2}+xyz}+yz.\sqrt{y^{2}z^{2}+xyz}+zx.\sqrt{z^{2}x^{2}+xyz}$$
$$\leqslant \frac{\sqrt2}{2}.\sqrt{xyz}.\sqrt{xy+yz+zx-xyz}$$
$$P^{2}=\frac{x^2y^2}{zx+zy}+\frac{y^2z^2}{xy+xz}+\frac{z^2x^2}{yx+yz}+2(\frac{x^2}{\sqrt{x+yz}}+\frac{y^2}{\sqrt{y+xz}}+\frac{z^2}{\sqrt{z+xy}})$$
Giải
Ta có: $P\leqslant \frac{\sqrt2}{2}$ $$\Leftrightarrow xy\sqrt{(xy+xz)(xy+yz)}+yz\sqrt{(yz+xy)(yz+zx)}+zx\sqrt{(zx+xy)(zx+yz)}$$
$$\leq \frac{\sqrt2}{2}\sqrt{(xy+yz)(yz+zx)(zx+xy)}$$
$$\Leftrightarrow xy.\sqrt{x^{2}y^{2}+xyz}+yz.\sqrt{y^{2}z^{2}+xyz}+zx.\sqrt{z^{2}x^{2}+xyz}$$
$$\leqslant \frac{\sqrt2}{2}.\sqrt{xyz}.\sqrt{xy+yz+zx-xyz}$$
$$P^{2}=\frac{x^2y^2}{zx+zy}+\frac{y^2z^2}{xy+xz}+\frac{z^2x^2}{yx+yz}+2(\frac{x^2}{\sqrt{x+yz}}+\frac{y^2}{\sqrt{y+xz}}+\frac{z^2}{\sqrt{z+xy}})$$
Nhận xét
Đăng nhận xét