Cho xin slot bài này :v

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:

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

Áp dụng Cauchy-Schwarz, ta có:

\(VT\ge\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{7}{ab+bc+ca}\)

\(VT\ge\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}+\dfrac{7}{\dfrac{\left(a+b+c\right)^2}{3}}=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{\dfrac{1}{3}}=9+21=30\)