Question

a.b>= 0 thoả mãn a+b+c = 1 CMR

\(\dfrac{ab}{c+1}+\dfrac{bc}{a+1}+\dfrac{ac}{b+1}< =\dfrac{1}{4}\)

Answer 1

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

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

\(\Rightarrow \frac{ab}{c+1}\leq \frac{1}{4}\left(\frac{ab}{c+a}+\frac{ab}{c+b}\right)\)

Tương tự:

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

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

Cộng theo vế các BĐT vừa thu được:

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

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$