Chủ đề này có 2 trả lời

[VNSTaipro]

Cho $0<a,b,c\leq 1$. Chứng minh:

$(1+\frac{1}{abc})(a+b+c)\geq 3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

 

[Hoang Tung 126]

Đặt $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$

Ta có :$a+b+c=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\geq 3\sqrt[3]{\frac{xyz}{xyz}}=3$

Mà :$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\geq \frac{y}{x}+\frac{z}{y}+\frac{x}{z}< = > x^2z+y^2x+z^2y\geq y^2z+x^2y+xz^2< = > xz(x-z)+y^2(x-z)-y(x-z)(x+z)\geq 0< = > (x-z)(xz+y^2-xy-yz)\geq 0< = > (x-z)(x(z-y)-y(z-y))\geq 0< = > (x-z)(x-y)(z-y)\geq 0$

[VNSTaipro]

Đặt $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$

Ta có :$a+b+c=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\geq 3\sqrt[3]{\frac{xyz}{xyz}}=3$

Mà :$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\geq \frac{y}{x}+\frac{z}{y}+\frac{x}{z}< = > x^2z+y^2x+z^2y\geq y^2z+x^2y+xz^2< = > xz(x-z)+y^2(x-z)-y(x-z)(x+z)\geq 0< = > (x-z)(xz+y^2-xy-yz)\geq 0< = > (x-z)(x(z-y)-y(z-y))\geq 0< = > (x-z)(x-y)(z-y)\geq 0$

Nếu đặt như vậy là tự cho $abc=1$ rồi bạn