1,Cho a,b>0. Chứng minh : $(\sqrt{a}+\sqrt{b})^{8}\geq 64ab(a+b)^{2}$
2, Cho x,y >0. Chứng minh: $\left ( \frac{x}{y} +\frac{y}{x}\right )^{32}\geq 8^{9}\left ( \frac{x^{2}}{y^{2}} +\frac{y^{2}}{x^{2}}\right )^{4}.\left ( \frac{x^{4}}{y^{4}} +\frac{y^{4}}{x^{4}}\right )$
1. Ta có :
$$(\sqrt{a}+\sqrt{b})^8=(a+b+2\sqrt{ab})^4\geq \left [ 2\sqrt{(a+b).2\sqrt{ab}} \right ]^4=64ab(a+b)^2$$
2. Áp dụng câu kết quả câu 1 ta có :
$$\left ( \frac{x}{y}+\frac{y}{x} \right )^8\geq 64\left ( \frac{x^2}{y^2}+\frac{y^2}{x^2} \right )^2$$
$$\Rightarrow \left ( \frac{x}{y}+\frac{y}{x} \right )^{32}\geq 8^8\left ( \frac{x^2}{y^2}+\frac{y^2}{x^2} \right )^8$$
Lại có
$$\left ( \frac{x^2}{y^2}+\frac{y^2}{x^2} \right )^4\geq \sqrt{64\left ( \frac{x^4}{y^4}+\frac{y^4}{x^4} \right )^2}=8\left ( \frac{x^4}{y^4}+\frac{y^4}{x^4} \right )$$
$$\Rightarrow \left ( \frac{x}{y}+\frac{y}{x} \right )^{32}\geq 8^8\left ( \frac{x^2}{y^2}+\frac{y^2}{x^2} \right )^4.8\left ( \frac{x^4}{y^4}+\frac{y^4}{x^4} \right )= 8^9\left ( \frac{x^{2}}{y^{2}} +\frac{y^{2}}{x^{2}}\right )^{4}\left ( \frac{x^{4}}{y^{4}} +\frac{y^{4}}{x^{4}}\right )$$