Question

cho a,b,c dương thỏa mãn ab+bc+ca=4. CMR: \(a+b+c\ge ab+bc+ca\)

Answer 1

Bạn post nhiều bài BĐT hay thật

Đặt \(\left(a;b;c\right)=\left(\dfrac{2x}{y+z};\dfrac{2y}{z+x};\dfrac{2z}{x+y}\right)\)

BĐT trở thành:

\(\sum_{cyc}\dfrac{x}{y+z}\ge\sum_{cyc}\dfrac{2xy}{\left(x+y\right)\left(x+z\right)}\)

Sử dụng AM-GM, ta có:

\(VP\le\sum_{cyc}xy\left[\dfrac{1}{\left(x+y\right)^2}+\dfrac{1}{\left(x+z\right)^2}\right]=\sum_{cyc}\dfrac{xy}{\left(z+x\right)^2}+\sum_{cyc}\dfrac{xy}{\left(y+z\right)^2}=\sum_{cyc}\dfrac{xy}{\left(y+z\right)^2}+\sum\dfrac{zx}{\left(y+z\right)^2}=\sum_{cyc}\dfrac{x}{y+z}=VT\)