Question

Cho a, b, c > 0 thỏa mãn: ab + bc + ca + abc ≤ 4. CMR: a² + b² + c² + a + b + c ≥ 2(ab+bc+ca)

Answer 1

Ta cần chứng minh

\(a+b+c\ge ab+bc+ca\)

do \(x^2+y^2+z^2\ge xy+yz+zx\)

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

\(\Rightarrow\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{x}{y+z}\ge2\left(\dfrac{xy}{\left(x+z\right)\left(y+z\right)}+\dfrac{yz}{\left(x+z\right)\left(x+y\right)}+\dfrac{zx}{\left(x+y\right)\left(y+z\right)}\right)\)

\(\Leftrightarrow x^3+y^3+z^3+3xyz\ge xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)\)

dấu ''='' khi \(a=b=c=1\) hoặc \(a=b=2,c=1\)