Question

1.Cho các số dương a,b. Chứng minh rằng \(\dfrac{1}{a}\)+\(\dfrac{1}{b}\)≥\(\dfrac{4}{a+b}\)

2. Cho a,b,c là các số thực không âm. Chứng minh rằng (a+b)(b+c)(c+a)≥8abc

Answer 1

1) xét hiệu

\(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{4}{a+b}\ge0\)

<=> \(\dfrac{b\left(a+b\right)}{ab\left(a+b\right)}+\dfrac{a\left(a+b\right)}{ab\left(a+b\right)}-\dfrac{4ab}{ab\left(a+b\right)}\ge0\)

=> b(a+b)+a(a+b)-4ab ≥ 0

<=> ab+b²+a²+ab-4ab ≥ 0

<=> a² -2ab+b² ≥ 0

<=> (a-b)² ≥ 0 (luôn đúng )

=> đpcm

Answer 2

2)Ta có:\(\left(a-b\right)^2\ge0\)

\(\Rightarrow a^2-2ab+b^2\ge0\)

\(\Rightarrow a^2+2ab+b^2-4ab\ge0\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\)

TT\(\Rightarrow\left(b+c\right)^2\ge4bc;\left(c+a\right)^2\ge4ca\)

\(\Rightarrow\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\ge64a^2b^2c^2\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)