Cho $0<a,b,c\leq 1$. Chứng minh:
$(1+\frac{1}{abc})(a+b+c)\geq 3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Cho $0<a,b,c\leq 1$. Chứng minh:
$(1+\frac{1}{abc})(a+b+c)\geq 3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
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 :$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\geq \frac{y}{x}+\frac{z}{y}+\frac{x}{z}< = > x^2z+y^2x+z^2y\geq y^2z+x^2y+xz^2< = > xz(x-z)+y^2(x-z)-y(x-z)(x+z)\geq 0< = > (x-z)(xz+y^2-xy-yz)\geq 0< = > (x-z)(x(z-y)-y(z-y))\geq 0< = > (x-z)(x-y)(z-y)\geq 0$
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
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