Toán x/1+y ² +y/+z ² +z/1+x ² ≥3/2 với x+y+z=3 07/09/2021 By Iris x/1+y ² +y/+z ² +z/1+x ² ≥3/2 với x+y+z=3
Giải thích các bước giải: $\dfrac{x}{{1 + {y^2}}} = x – \dfrac{{x{y^2}}}{{1 + {y^2}}} \ge x- \dfrac{{x{y^2}}}{{2y}} = x – \dfrac{{xy}}{2}\\ \rightarrow \text{Tương tự: }\\ \rightarrow \dfrac{y}{{1 + {z^2}}}\ge y – \dfrac{{yz}}{2}\\ \quad \dfrac{z}{{1 + {x^2}}}\ge z – \dfrac{{zx}}{2}\\ \Rightarrow \dfrac{x}{1+y^2}+\dfrac{y}{1+z^2}+\dfrac{z}{1+x^2}\ge (x+y+z)-\dfrac{xy+yz+zx}{2}\ge \dfrac{3}{2}$ Trả lời
Giải thích các bước giải:
$\dfrac{x}{{1 + {y^2}}} = x – \dfrac{{x{y^2}}}{{1 + {y^2}}} \ge x- \dfrac{{x{y^2}}}{{2y}} = x – \dfrac{{xy}}{2}\\
\rightarrow \text{Tương tự: }\\
\rightarrow \dfrac{y}{{1 + {z^2}}}\ge y – \dfrac{{yz}}{2}\\
\quad \dfrac{z}{{1 + {x^2}}}\ge z – \dfrac{{zx}}{2}\\
\Rightarrow \dfrac{x}{1+y^2}+\dfrac{y}{1+z^2}+\dfrac{z}{1+x^2}\ge (x+y+z)-\dfrac{xy+yz+zx}{2}\ge \dfrac{3}{2}$