Question

1) chứng minh:

a) 4x² - xy +y² ≥ 0

b) a² + b² + 2c² ≥ 2c(a+b)

c) a⁴ + b⁴ + c⁴ + d⁴ ≥ 4abcd

Answer 1

1)a)\(4x^2-xy+y^2\ge0\)

\(\Leftrightarrow\left(\dfrac{1}{4}x^2-xy+y^2\right)+\dfrac{15}{4}x^2\ge0\)

\(\Leftrightarrow\left(\dfrac{1}{2}x-y\right)^2+\dfrac{15}{4}x^2\ge0\)(luôn đúng)

b)\(a^2+b^2+2c^2\ge2c\left(a+b\right)\)

\(\Leftrightarrow a^2+b^2+2c^2-2ac-2bc\ge0\)

\(\Leftrightarrow\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2\ge0\)(luôn đúng)

c)Ta có:\(\left(a^2-b^2\right)\ge0\)

\(\Rightarrow a^4+b^4\ge2a^2b^2\)(1)

TT\(\Rightarrow c^4+d^4\ge2c^2d^2\)(2)

\(2a^2b^2+2c^2d^2\ge4abcd\left(3\right)\)

Từ (1)(2)(3)=>đpcm