Question

[Đào Linh]

cho a;b;c>0. chứng minh rằng \(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge a\sqrt{ac}+b\sqrt{ba}+c\sqrt{cb}\)

Answer 1

[Đặng Anh Thư]

ta có : \(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}=\dfrac{a^3}{b}+bc+\dfrac{b^3}{c}+ca+\dfrac{c^3}{a}+ab-\left(ac+bc+ab\right)\)

\(=\dfrac{a^3}{b}+bc+\dfrac{b^3}{c}+ca+\dfrac{c^3}{a}+ab-\left(\dfrac{ab}{2}+\dfrac{bc}{2}+\dfrac{ab}{2}+\dfrac{ac}{2}+\dfrac{bc}{2}+\dfrac{ac}{2}\right)\)

\(\ge2.\sqrt{\dfrac{a^3}{b}.bc}+2\sqrt{\dfrac{b^3}{c}.ca}+2\sqrt{\dfrac{c^3}{a}.ab}-2\sqrt{\dfrac{ab.bc}{4}}-2\sqrt{\dfrac{ab.ac}{4}}-2\sqrt{\dfrac{bc.ac}{4}}\)

\(\ge2a\sqrt{ac}+2b\sqrt{ba}+2c\sqrt{cb}-b\sqrt{ac}-a\sqrt{bc}-c\sqrt{ab}=a\sqrt{ac}+b\sqrt{ba}+c\sqrt{cb}\left(ĐPCM\right)\)

Answer 2

[Neet]

Áp dụng BĐT cauchy-schwarz:

\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\)

BĐT cần chứng minh tương đương :

\(\left(a+b+c\right)^2\ge3\left(\sqrt{a^3c}+\sqrt{b^3a}+\sqrt{c^3b}\right)\)

Thật vậy, Áp dụng BĐT \(\left(X+Y+Z\right)^2\ge3\left(XY+YZ+ZX\right)\)

Với \(\left\{{}\begin{matrix}X=a+\sqrt{bc}-\sqrt{ac}\\Y=b+\sqrt{ac}-\sqrt{ab}\\Z=c+\sqrt{ab}-\sqrt{bc}\end{matrix}\right.\) ta có ngay ĐPCM. ( mất chút time khai triển)

Dấu = xảy ra khi X=Y=Z hay a=b=c