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$\Leftrightarrow x=y=z=t=\frac{1}{2}$

BĐT

1/ Cho $a, b, c, d > 0 ; ab \leq 1 ; cd \leq 1$ . CMR :$\frac{1}{1+a}$ + $\frac{1}{1+b}$ + $\frac{1}{1+c}$ + $\frac{1}{1+d}$ $\leq$ $\frac{4}{1+\sqrt[4]{abcd}}$ 2/ Cho $a, b, c, d > 0$ thỏa $abcd = 1$ . CMR :$\frac{1}{(1+a)^2}$ + $\frac{1}{(1+b)^2}$ + $\frac{1}{(1+c)^2}$ + $\frac{1}{(1+d)^2}$ $\geq 1$

Bất đẳng thức

BĐT

1/ Cho $a, b, c, d > 0 ; ab \leq 1 ; cd \leq 1$ . CMR :$\frac{1}{1+a}$ + $\frac{1}{1+b}$ + $\frac{1}{1+c}$ + $\frac{1}{1+d}$ $\leq$ $\frac{4}{1+\sqrt[4]{abcd}}$

Bất đẳng thức

PT

Giải PT : $\left| {x-3} \right|x^{16} + \left| {x-4} \right|x^{17} = 1$

Phương trình chứa dấu...

PT

Giải PT : $\left| {x-3} \right|^{16} + \left| {x-4} \right|^{17} = 1$

Phương trình chứa dấu...