Đáp án:
Giải thích các bước giải:
`1/((b+c)a^3)+1/((a+c)b^3)+1/((b+a)c^3)`
`=(1:a^2)/((b+c)a^3:a^2)+(1:b^2)/((a+c)b^3:b^2)+(1:c^2)/((b+a)c^3:c^2)`
$\dfrac{\dfrac{1}{a^2}}{(b+c)a}+$ $\dfrac{\dfrac{1}{b^2}}{(a+c)b}+$ $\dfrac{\dfrac{1}{c^2}}{(b+a)c}$
Áp dụng bất đẳng thức `AM-GM`
`=>`$\dfrac{\dfrac{1}{a^2}}{(b+c)a}+$ $\dfrac{\dfrac{1}{b^2}}{(a+c)b}+$ $\dfrac{\dfrac{1}{c^2}}{(b+a)c}$ $\geq$ $\dfrac{(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^2}{(b+c)a+(a+c)b+(b+a)c}$
`=>`$\dfrac{\dfrac{1}{a^2}}{(b+c)a}+$ $\dfrac{\dfrac{1}{b^2}}{(a+c)b}+$ $\dfrac{\dfrac{1}{c^2}}{(b+a)c}$$\geq$ $\dfrac{\dfrac{ab+ac+bc}{abc}}{2(ab+ac+bc)}$
`=>`$\dfrac{\dfrac{1}{a^2}}{(b+c)a}+$ $\dfrac{\dfrac{1}{b^2}}{(a+c)b}+$ $\dfrac{\dfrac{1}{c^2}}{(b+a)c}$$\geq$ `(ab+ac+bc)/2`
Có:`ab+ac+bc>=3.`$\sqrt[3]{ab.bc.ac}$
`=>ab+ac+bc>=3`
`=>(ab+ac+bc)/2 >=3/2`
`=>`$\dfrac{\dfrac{1}{a^2}}{(b+c)a}+$ $\dfrac{\dfrac{1}{b^2}}{(a+c)b}+$ $\dfrac{\dfrac{1}{c^2}}{(b+a)c}$`>=3/2`
Dấu `=` xảy ra `<=>a=b=c=1`
