Question

Cho a, b, c > 0. CMR:

a) \(\frac{a^2}{a+b}\) + \(\frac{b^2}{b+c}\) + \(\frac{c^2}{c+a}\) ≥ \(\frac{ab}{a+b}\) + \(\frac{bc}{b+c}\) + \(\frac{ca}{c+a}\)

b) a²b²(a² + b²) ≤ 2 với a + b = 2 và a; b > 0

Answer 1

a) Ta có: \(\frac{a^2}{a+b}-\frac{b^2}{a+b}+\frac{b^2}{b+c}-\frac{c^2}{b+c}+\frac{c^2}{c+a}-\frac{a^2}{c+a}\) \(=a-b+b-c+c-a=0\)

\(\Rightarrow\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}=\frac{b^2}{a+b}+\frac{c^2}{b+c}+\frac{a^2}{c+a}\)

\(\Rightarrow2\left(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\right)=\frac{a^2}{a+b}+\frac{b^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{b+c}+\frac{c^2}{c+a}+\frac{a^2}{c+a}\)\(\ge\frac{2ab}{a+b}+\frac{2bc}{b+c}+\frac{2ca}{c+a}\)

\(\Rightarrowđpcm\)

Dấu "=" \(\Leftrightarrow a=b=c\)

b) \(a^2b^2\left(a^2+b^2\right)=\frac{1}{2}\cdot ab\cdot2ab\cdot\left(a^2+b^2\right)\le\frac{1}{2}\cdot\frac{\left(a+b\right)^2}{4}\cdot\frac{\left(2ab+a^2+b^2\right)^2}{4}=2\)

Dấu "=" \(\Leftrightarrow a=b=1\)