Question

Cho các số thực dương a , b , c thỏa mãn \(a+b+c=3\)

Chứng minh rằng \(\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)

Answer 1

Áp dụng bất đẳng thức cộng mẫu số

\(\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)

\(\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)

Chứng minh rằng \(\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)

\(\Leftrightarrow18\ge3\left(3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}\right)\)

\(\Leftrightarrow18\ge9+3\sqrt[3]{bc}+3\sqrt[3]{ca}+3\sqrt[3]{ab}\)

\(\Leftrightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\)

Áp dụng bất đẳng thức Cauchy cho 3 bộ số thực không âm

\(\Rightarrow\left\{{}\begin{matrix}a+b+1\ge3\sqrt[3]{ab}\\b+c+1\ge3\sqrt[3]{bc}\\c+a+1\ge3\sqrt[3]{ca}\end{matrix}\right.\)

\(\Rightarrow2\left(a+b+c\right)+3\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\)

\(\Rightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\) ( đpcm )

Vì \(\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)

Mà \(\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)

\(\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)( đpcm )

Answer 2

Áp dụng BĐT AM-GM và Cauchy-Schwarz ta có:

\(\sum\frac{a^2}{a+\sqrt[3]{bc}}\geq\sum\frac{a^2}{a+\frac{b+c+1}{3}}=\sum\frac{9a^2}{3(3a+b+c)+a+b+c}\)

\(=\sum\frac{9a^2}{10a+4b+4c}\geq\frac{9(a+b+c)^2}{(10a+4b+4c)}=\frac{9(a+b+c)^2}{18(a+b+c)}=\frac{3}{2}\)

Answer 3

Với x,y,zx,y,z không âm thỏa mãn x≥y≥z≥0x≥y≥z≥0 thì ta có các chú ý sau:

1.xy+yz+xzx2+xz+z2≥y+zx+z1.xy+yz+xzx2+xz+z2≥y+zx+z

2.xy+yz+xzy2+yz+z2≥x+zy+z2.xy+yz+xzy2+yz+z2≥x+zy+z

3.xy+yz+xzx2+xy+y2≥(x+z)(y+z)(x+z)2+(x+z)(y+z)+(y+z)23.xy+yz+xzx2+xy+y2≥(x+z)(y+z)(x+z)2+(x+z)(y+z)+(y+z)2

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Với những công cụ hỗ trợ này, ta có thể xử đẹp bài toán sau:

xy+yz+xzx2+xy+y2−−−−−−−−−−−√+xy+yz+xzy2+yz+z2−−−−−−−−−−−√+xy+yz+xzz2+zx+z2−−−−−−−−−−−√≥2+13–√