Ch0 a,b,c >0 , `ab+bc+ca=3` Min P= `(a^3)/((1+b)(1+c))+(b^3)/((1+c)(1+a))+(c^3)/((1+a)(1+b))=?`

Đáp án:

 $P_{min}=\dfrac{3}{4}$

Giải thích các bước giải:

 $\dfrac{a^3}{(1+b)(1+c)}+\dfrac{1+b}{8}+\dfrac{1+c}{8} \geq 3\sqrt[3]{\dfrac{a^3(1+b)(1+c)}{64(1+b)(1+c)}}=\dfrac{3a}{4}$

Tương tự:

$\dfrac{b^3}{(1+c)(1+a)}+\dfrac{1+c}{8}+\dfrac{1+a}{8} \geq \dfrac{3b}{4}$

$\dfrac{c^3}{(1+a)(1+b)}+\dfrac{1+a}{8}+\dfrac{1+b}{8} \geq \dfrac{3c}{4}$

Cộng vế:

$P+\dfrac{6+2(a+b+c)}{8} \geq \dfrac{3(a+b+c)}{4}$

$⇒P \geq \dfrac{a+b+c}{2}-\dfrac{3}{4} \geq \dfrac{\sqrt{3(ab+bc+ca)}}{2}-\dfrac{3}{4}=\dfrac{3}{4}$

$P_{min}=\dfrac{3}{4}$ khi $a=b=c=1$