Quay lại đáp án

Theo Cauchy-Schwarz$2=\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{t} \geq \frac{16}{x+y+z+t}$$\Rightarrow x+y+z+t \geq 2$$\frac{x^3}{x^2+y^2}=x-\frac{xy^2}{x^2+y^2}$Theo BĐT Cauchy: $x^2+y^2\geq2xy$$\Rightarrow \frac{xy^2}{x^2+y^2}\leq\frac{xy^2}{2xy}=\frac{y}{2}$Tương tự với $ y,z,t$, ta có :$\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+t^2}+\frac{t^3}{t^2+x^2}=(x-\frac{xy^2}{x^2+y^2})+(y-\frac{yz^2}{y^2+z^2})+(z-\frac{zt^2}{z^2+t^2})+(t-\frac{tx^2}{t^2+x^2})\geq x-\frac{y}{2}+y-\frac{z}{2}+z-\frac{t}{2}+t-\frac{x}{2}=\frac{x}{2}+\frac{y}{2}+\frac{z}{2}+\frac{t}{2}\geq1$Dấu $"="$ xảy ra $\Leftrightarrow x=y=z=t=\frac{1}{2}$

Theo Cauchy-Schwarz$2=\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{t} \geq \frac{16}{x+y+z+t}$$\Rightarrow x+y+z+t \geq 8$$\frac{x^3}{x^2+y^2}=x-\frac{xy^2}{x^2+y^2}$Theo BĐT Cauchy: $x^2+y^2\geq2xy$$\Rightarrow \frac{xy^2}{x^2+y^2}\leq\frac{xy^2}{2xy}=\frac{y}{2}$Tương tự với $ y,z,t$, ta có :$\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+t^2}+\frac{t^3}{t^2+x^2}=(x-\frac{xy^2}{x^2+y^2})+(y-\frac{yz^2}{y^2+z^2})+(z-\frac{zt^2}{z^2+t^2})+(t-\frac{tx^2}{t^2+x^2})\geq x-\frac{y}{2}+y-\frac{z}{2}+z-\frac{t}{2}+t-\frac{x}{2}=\frac{x}{2}+\frac{y}{2}+\frac{z}{2}+\frac{t}{2}\geq4$Dấu $"="$ xảy ra $\Leftrightarrow x=y=z=t=2$

Theo Cauchy-Schwarz$2=\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{t} \geq \frac{16}{x+y+z+t}$$\Rightarrow x+y+z+t \geq 2$$\frac{x^3}{x^2+y^2}=x-\frac{xy^2}{x^2+y^2}$Theo BĐT Cauchy: $x^2+y^2\geq2xy$$\Rightarrow \frac{xy^2}{x^2+y^2}\leq\frac{xy^2}{2xy}=\frac{y}{2}$Tương tự với $ y,z,t$, ta có :$\frac{x^2}{x^2+y^2}+\frac{y^2}{y^2+z^2}+\frac{z^2}{z^2+t^2}+\frac{t^2}{t^2+x^2}=(x-\frac{xy^2}{x^2+y^2})+(y-\frac{yz^2}{y^2+z^2})+(z-\frac{zt^2}{z^2+t^2})+(t-\frac{tx^2}{t^2+x^2})\geq x-\frac{y}{2}+y-\frac{z}{2}+z-\frac{t}{2}+t-\frac{x}{2}=\frac{x}{2}+\frac{y}{2}+\frac{z}{2}+\frac{t}{2}\geq1$Dấu $"="$ xảy ra $\Leftrightarrow x=y=z=t=\frac{1}{2}$

Theo Cauchy-Schwarz$2=\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{t} \geq \frac{16}{x+y+z+t}$$\Rightarrow x+y+z+t \geq 2$$\frac{x^3}{x^2+y^2}=x-\frac{xy^2}{x^2+y^2}$Theo BĐT Cauchy: $x^2+y^2\geq2xy$$\Rightarrow \frac{xy^2}{x^2+y^2}\leq\frac{xy^2}{2xy}=\frac{y}{2}$Tương tự với $ y,z,t$, ta có :$\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+t^2}+\frac{t^3}{t^2+x^2}=(x-\frac{xy^2}{x^2+y^2})+(y-\frac{yz^2}{y^2+z^2})+(z-\frac{zt^2}{z^2+t^2})+(t-\frac{tx^2}{t^2+x^2})\geq x-\frac{y}{2}+y-\frac{z}{2}+z-\frac{t}{2}+t-\frac{x}{2}=\frac{x}{2}+\frac{y}{2}+\frac{z}{2}+\frac{t}{2}\geq1$Dấu $"="$ xảy ra $\Leftrightarrow x=y=z=t=\frac{1}{2}$