Cho a + b + c = 2. Tìm giá trị lớn nhất của: P=ab/ căn ab+2c + bc/ căn bc + 2a + ca / căn ca+2b

Trả lời

Theo BĐT Cauchy-Schwarz ta có:

$\frac{ab}{\sqrt{ab+2c}}=\frac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\frac{ab}{\sqrt{ab+ac+bc+c^2}}=\frac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le \frac{1}{2}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)$

Tương tự ta cũng có:

$\frac{bc}{\sqrt{bc+2a}}\le \frac{1}{2}\left(\frac{bc}{a+b}+\frac{bc}{a+c}\right)$

 $\frac{ca}{\sqrt{ca+2b}}\le \frac{1}{2}\left(\frac{ca}{a+b}+\frac{ca}{b+c}\right)$

Cộng theo vế 3 BĐT trên ta có:

 $P\le \frac{1}{2}\left(\frac{ab+bc}{a+c}+\frac{bc+ca}{a+b}+\frac{ab+ca}{b+c}\right)$

$=\frac{1}{2}\left(\frac{b\left(a+c\right)}{a+c}+\frac{c\left(a+b\right)}{a+b}+\frac{a\left(b+c\right)}{b+c}\right)=\frac{1}{2}\left(a+b+c\right)=\frac{1}{2}\cdot 2=1.$

Dấu “ = ” xảy ra khi  $a=b=c=\frac{2}{3}.$