Cho a,b,c là các số thực dương thỏa mãn a+b+c=1. Tìm giá trị lớn nhất của: P = $\frac{ab}{c+1}$ + $\frac{bc}{a+1}$ + $\frac{ac}{b+1}$

Đáp án: $P\le \dfrac14$

Giải thích các bước giải:

Ta có:

$P=\dfrac{ab}{c+1}+\dfrac{bc}{a+1}+\dfrac{ac}{b+1}$

$\to P=\dfrac{ab}{c+a+b+c}+\dfrac{bc}{a+a+b+c}+\dfrac{ac}{b+a+b+c}$

$\to P=\dfrac{ab}{(c+a)+(b+c)}+\dfrac{bc}{(a+b)+(c+a)}+\dfrac{ac}{(a+b)+(b+c)}$

$\to P=ab\cdot\dfrac{1}{(c+a)+(b+c)}+bc\cdot\dfrac{1}{(a+b)+(c+a)}+ca\cdot\dfrac{1}{(a+b)+(b+c)}$

$\to P\le ab\cdot \dfrac14(\dfrac1{c+a}+\dfrac1{b+c})+bc\cdot \dfrac14(\dfrac1{a+b}+\dfrac1{c+a})+ca\cdot \dfrac14(\dfrac1{a+b}+\dfrac1{b+c})$

$\to P\le \dfrac14(\dfrac{ab}{c+a}+\dfrac{ab}{b+c}+\dfrac{bc}{a+b}+\dfrac{bc}{c+a}+\dfrac{ca}{a+b}+\dfrac{ca}{b+c})$

$\to P\le \dfrac14((\dfrac{ab}{c+a}+\dfrac{bc}{c+a})+(\dfrac{ab}{b+c}+\dfrac{ca}{b+c})+(\dfrac{bc}{a+b}+\dfrac{ca}{a+b})$

$\to P\le \dfrac14(b+a+c)$

$\to P\le \dfrac14$

Dấu = xảy ra khi $a=b=c=\dfrac12$