Quay lại đáp án

câu 9gt $\Leftrightarrow 5x^{2}+5(y^{2}+z^{2})-9x(y+z)-18yz =0$ $yz\leq \frac{1}{4} (y+z)^{2} ;y^{2}+z^{2} \geq \frac{1}{2}(y+z)^{2}$ $\Rightarrow 18yz-5(y^{2}+z^{2})\leq 2(y+z)^{2} \Rightarrow 5x^{2}-9x(y+z)\leq 2(y+z)^{2}$$\Leftrightarrow (x-2(y+z))(5x+y+z)\leq0$ $\Rightarrow x\leq 2(y+z)$ P=$\frac{x}{y^{2}+z^{2}} -\frac{1}{(x+y+z)^{3}} \leq \frac{2x}{(y+z)^{2}}-\frac{1}{(x+y+z)^{3}}$ $\leq \frac{4}{y+z}-\frac{1}{27(y+z)^{3}}$ Đặt $\frac{1}{y+z}=t$ $\Rightarrow P\leq 4t-\frac{1}{27}t^{3}$ Đánh giá kiểu j để P $\leq 16$dấu '="$\Leftrightarrow x=\frac{1}{3};y=z=\frac{1}{12}$

câu 9gt $\Leftrightarrow 5x^{2}+5(y^{2}+z^{2})-9x(y+z)-18yz =0$ $yz\leq \frac{1}{4} (y+z)^{2} ;y^{2}+z^{2} \geq \frac{1}{2}(y+z)^{2}$ $\Rightarrow 18yz-5(y^{2}+z^{2})\leq 2(y+z)^{2} \Rightarrow 5x^{2}-9x(y+z)\leq 2(y+z)^{2}$$\Leftrightarrow (x-2(y+z))(5x+y+z)\leq0$ $\Rightarrow x\leq 2(y+z)$ P=$\frac{x}{y^{2}+z^{2}} -\frac{1}{(x+y+z)^{3}} \leq \frac{2x}{(y+z)^{2}}-\frac{1}{(x+y+z)^{3}}$ $\leq \frac{4}{y+z}-\frac{1}{27(y+z)^{3}}$ Đặt $\frac{1}{y+z}=t$ $\Rightarrow P\leq 4t-\frac{1}{27}t^{3}$Ta sẽ chứng minh $4t-\frac{1}{27}t^{3} \leq 16$$\Leftrightarrow \frac{(t-6)^2(t+12)}{27} \ge0$ (đúng $\forall t >0$)$\Rightarrow P \le 16$dấu '="$\Leftrightarrow x=\frac{1}{3};y=z=\frac{1}{12}$

câu 9gt $\Leftrightarrow 5x^{2}+5(y^{2}+z^{2})-9x(y+z)-18yz =0$ $yz\leq \frac{1}{4} (y+z)^{2} ;y^{2}+z^{2} \geq \frac{1}{2}(y+z)^{2}$ $\Rightarrow 18yz-5(y^{2}+z^{2})\leq 2(y+z)^{2} \Rightarrow 5x^{2}-9x(y+z)\leq 2(y+z)^{2}$$\Leftrightarrow (x-2(y+z))(5x+y+z)\leq0 \Rightarrowx\leq 2(y+z)$ P=$\frac{x}{y^{2}+z^{2}} -\frac{1}{(x+y+z)^{3}} \leq \frac{2x}{(y+z)^{2}}-\frac{1}{(x+y+z)^{3}}$ $\leq \frac{4}{y+z}-\frac{1}{27(y+z)^{3}}$ Đặt $\frac{1}{y+z}=t$ $\Rightarrow P\leq 4t-\frac{1}{27}t^{3}$ Đánh giá kiểu j để P $\leq 16$dấu '=" $\Leftrightarrow x=\frac{1}{3};y=z=\frac{1}{12}$

câu 9gt $\Leftrightarrow 5x^{2}+5(y^{2}+z^{2})-9x(y+z)-18yz =0$ $yz\leq \frac{1}{4} (y+z)^{2} ;y^{2}+z^{2} \geq \frac{1}{2}(y+z)^{2}$ $\Rightarrow 18yz-5(y^{2}+z^{2})\leq 2(y+z)^{2} \Rightarrow 5x^{2}-9x(y+z)\leq 2(y+z)^{2}$$\Leftrightarrow (x-2(y+z))(5x+y+z)\leq0$ $\Rightarrow x\leq 2(y+z)$P=$\frac{x}{y^{2}+z^{2}} -\frac{1}{(x+y+z)^{3}} \leq \frac{2x}{(y+z)^{2}}-\frac{1}{(x+y+z)^{3}}\leq \frac{4}{y+z}-\frac{1}{27(y+z)^{3}}$Đặt$\frac{1}{y+z}=t\Rightarrow P\leq 4t-\frac{1}{27}t^{3}$Đánh giá kiểu j để P$\leq 16$dấu '="$\Leftrightarrow x=\frac{1}{3};y=z=\frac{1}{12}$

câu 9gt $\Leftrightarrow 5x^{2}+5(y^{2}+z^{2})-9x(y+z)-18yz =0$ $yz\leq \frac{1}{4} (y+z)^{2} ;y^{2}+z^{2} \geq \frac{1}{2}(y+z)^{2}$ $\Rightarrow 18yz-5(y^{2}+z^{2})\leq 2(y+z)^{2} \Rightarrow 5x^{2}-9x(y+z)\leq 2(y+z)^{2}$$\Leftrightarrow (x-2(y+z)(5x+y+z)\leq0 \Rightarrowx\leq 2(y+z)$ P=$\frac{x}{y^{2}+z^{2}} -\frac{1}{(x+y+z)^{3}} \leq \frac{2x}{(y+z)^{2}}-\frac{1}{(x+y+z)^{3}}$ $\leq \frac{4}{y+z}-\frac{1}{27(y+z)^{3}}$ Đặt $\frac{1}{y+z}=t$ $\Rightarrow P\leq 4t-\frac{1}{27}t^{3}$ Đánh giá kiểu j để P $\leq 16$dấu '=" $\Leftrightarrow x=\frac{1}{3};y=z=\frac{1}{12}$

câu 9gt $\Leftrightarrow 5x^{2}+5(y^{2}+z^{2})-9x(y+z)-18yz =0$ $yz\leq \frac{1}{4} (y+z)^{2} ;y^{2}+z^{2} \geq \frac{1}{2}(y+z)^{2}$ $\Rightarrow 18yz-5(y^{2}+z^{2})\leq 2(y+z)^{2} \Rightarrow 5x^{2}-9x(y+z)\leq 2(y+z)^{2}$$\Leftrightarrow (x-2(y+z))(5x+y+z)\leq0 \Rightarrowx\leq 2(y+z)$P=$\frac{x}{y^{2}+z^{2}} -\frac{1}{(x+y+z)^{3}} \leq \frac{2x}{(y+z)^{2}}-\frac{1}{(x+y+z)^{3}}\leq \frac{4}{y+z}-\frac{1}{27(y+z)^{3}}$Đặt$\frac{1}{y+z}=t\Rightarrow P\leq 4t-\frac{1}{27}t^{3}$Đánh giá kiểu j để P$\leq 16$dấu '="$\Leftrightarrow x=\frac{1}{3};y=z=\frac{1}{12}$