Question

Cho a, b, c > 0 . CMR:

A= \(\dfrac{1}{a^3+b^3+abc}+\dfrac{1}{b^3+c^3+abc}+\dfrac{1}{c^3+a^3+abc}\le\dfrac{1}{abc}\)

Answer 1

Ta chứng minh BĐT phụ \(x^3+y^3\ge xy\left(x+y\right)\left(x>0,y>0\right)\Leftrightarrow x^2-xy+y^2\ge xy\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)

Áp dụng BĐT, ta được \(A=\dfrac{1}{a^3+b^3+abc}+\dfrac{1}{b^3+c^3+abc}+\dfrac{1}{c^3+a^3+abc}\le\dfrac{1}{ab\left(a+b\right)+abc}+\dfrac{1}{bc\left(b+c\right)+abc}+\dfrac{1}{ca\left(c+a\right)+abc}=\dfrac{1}{ab\left(a+b+c\right)}+\dfrac{1}{bc\left(a+b+c\right)}+\dfrac{1}{ca\left(a+b+c\right)}=\dfrac{1}{a+b+c}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\dfrac{1}{a+b+c}.\dfrac{a+b+c}{abc}=\dfrac{1}{abc}\left(đpcm\right)\)

Dấu = xảy ra <=> a = b = c.