Question

a, Cho x,y là các số thực dương bất kỳ. Chứng minh \(\frac{1}{x+y}\) ≤ \(\frac{1}{4}\) ( \(\frac{1}{x}\) + \(\frac{1}{y}\) )

b, Cho a,b và c là các số thực không âm thỏa mãn a + b + c = 1

Chứng minh rằng \(\frac{ab}{c+1}\) + \(\frac{bc}{a+1}\) + \(\frac{ca}{b+1}\) ≤ \(\frac{1}{4}\)

Answer 1

a) \(\frac{1}{x}+\frac{1}{y}\ge\frac{\left(1+1\right)^2}{x+y}=\frac{4}{x+y}\)

\(\Leftrightarrow\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)

b)

Ta có

\(\frac{ab}{c+1}=\frac{ab}{a+b}=\frac{ab}{\left(a+c\right)+\left(b+c\right)}\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\)

\(\frac{bc}{a+1}=\frac{bc}{\left(a+b\right)+\left(a+c\right)}\le\frac{bc}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\)

\(\frac{ac}{b+1}=\frac{ac}{\left(a+b\right)+\left(b+c\right)}\le\frac{ac}{4}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)\)

\(\Leftrightarrow\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ac}{b+1}\le\frac{ab}{4\left(a+c\right)}+\frac{ab}{4\left(b+c\right)}+\frac{bc}{4\left(a+b\right)}+\frac{bc}{4\left(a+c\right)}+\frac{ac}{4\left(A+b\right)}+\frac{ac}{4\left(b+c\right)}\)

\(=\frac{ab+bc}{4\left(a+c\right)}+\frac{ab+ac}{4\left(b+c\right)}+\frac{bc+ac}{4\left(a+b\right)}=\frac{1}{4}\left(\frac{b\left(a+c\right)}{a+c}\right)+\frac{1}{4}\left(\frac{a\left(b+c\right)}{b+c}\right)+\frac{c\left(a+b\right)}{a+b}\)

\(=\frac{a+b+c}{4}=\frac{1}{4}\)