Có:
[tex](\frac{\sqrt{y+z}}{x}+\frac{\sqrt{z+ x}}{y}+\frac{\sqrt{x+y}}{z}).\sqrt{(y+z)(z+x)(x+y)}\\=\frac{(y+z).\sqrt{(z+x)(x+y})}{x}+\frac{(z+x).\sqrt{(x+y)(y+z)}}{y}+\frac{(x+y).\sqrt{(x+z)(y+z)}}{z}[/tex]
Áp dụng BĐT Cauchy-Schwarz [tex](a+b)(a+c)\geq (a+\sqrt{bc})^2\rightarrow \sqrt{(a+b)(a+c)}\geq a+\sqrt{bc}[/tex] Có
[tex]\frac{(y+z).\sqrt{(z+x)(x+y})}{x}+\frac{(z+x).\sqrt{(x+y)(y+z)}}{y}+\frac{(x+y).\sqrt{(x+z)(y+z)}}{z}\\\geq \frac{(y+z).(x+\sqrt{yz})}{x}+\frac{(z+x).(y+\sqrt{xz})}{y}+\frac{(x+y)(z+\sqrt{xy})}{z}\\=(y+z+\frac{(y+z)\sqrt{zy}}{x})+(y+x+\frac{(y+x)\sqrt{xy}}{z})+(x+z+\frac{(x+z)\sqrt{xz}}{y})\\=2(x+y+z)+\frac{(y+z)\sqrt{zy}}{x}+\frac{(y+x)\sqrt{xy}}{z}+\frac{(x+z)\sqrt{xz}}{y}\\=2(x+y+z)+\frac{(y+z)\sqrt{zy}}{x}+\frac{(y+x)\sqrt{xy}}{z}+\frac{(x+z)\sqrt{xz}}{y}[/tex]
Áp dụng BĐT Cauchy [tex]a+b\geq 2\sqrt{ab}[/tex] có:
[tex]2(x+y+z)+\frac{(y+z)\sqrt{zy}}{x}+\frac{(y+x)\sqrt{xy}}{z}+\frac{(x+z)\sqrt{xz}}{y}\\\geq 2(x+y+z)+\frac{2\sqrt{yz}.\sqrt{yz}}{x}+\frac{2\sqrt{yx}.\sqrt{yx}}{z}+\frac{2\sqrt{xz}.\sqrt{xz}}{y}\\=2(x+y+z)+\frac{2yz}{x}+\frac{2xy}{z}+\frac{2zx}{y}\\=2(x+y+z)+\frac{2y^2z^2}{xyz}+\frac{2x^2y^2}{xyz}+\frac{2z^2x^2}{xyz}\\=2(x+y+z)+\frac{2(x^2y^2+y^2z^2+z^2x^2)}{xyz}[/tex]
Áp dụng BĐT phụ [tex]a^2+b^2+c^2\geq ab+bc+ca[/tex] có
[tex]2(x+y+z)+\frac{2(x^2y^2+y^2z^2+z^2x^2)}{xyz}\\\geq 2(x+y+z)+\frac{2(x^2yz+y^2xz+z^2xy)}{xyz}\\=2(x+y+z)+\frac{2xyz(x+y+z)}{xyz}\\=2(x+y+z)+2(x+y+z)\\=4(x+y+z)[/tex]
Suy ra
[tex](\frac{\sqrt{y+z}}{x}+\frac{\sqrt{z+ x}}{y}+\frac{\sqrt{x+y}}{z}).\sqrt{(y+z)(z+x)(x+y)} \geq 4(x+y+z)\\\rightarrow dpcm[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
