Cho 3 số thực dương a b c thỏa mãn a+b+c=3. tìm giá trị nhỏ nhất của biểu thức P= $\frac{ab}{ c^{2} (a+b) }$ + $\frac{ac}{ b^{2} (a+c) }$ $\frac{bc}{ a^{2} (b+c) }$

Đáp án: $P\ge \dfrac32$

Giải thích các bước giải:

Ta có :
$P=\dfrac{ab}{c^2\left(a+b\right)}+\dfrac{ac}{b^2\left(a+c\right)}+\dfrac{bc}{a^2\left(b+c\right)}$

$\to P=\dfrac{\left(\dfrac1c\right)^2}{\dfrac1a+\dfrac1b}+\dfrac{\left(\dfrac1b\right)^2}{\dfrac1a+\dfrac1c}+\dfrac{\left(\dfrac1a\right)^2}{\dfrac1c+\dfrac1b}$

$\to P\ge \dfrac{\left(\dfrac1c+\dfrac1b+\dfrac1a\right)^2}{\dfrac1a+\dfrac1b+\dfrac1a+\dfrac1c+\dfrac1c+\dfrac1b}$

$\to P\ge\dfrac{\left(\dfrac1a+\dfrac1b+\dfrac1c\right)^2}{2\left(\dfrac1a+\dfrac1b+\dfrac1c\right)}$

$\to P\ge \dfrac12\left(\dfrac1a+\dfrac1b+\dfrac1c\right)$

$\to P\ge \dfrac12.\dfrac9{a+b+c}$

$\to P\ge \dfrac32$

Dấu = xảy ra khi $a=b=c=1$