Question

[Trịnh Bá Vương Toàn]

- Cho a,b,c là các số thực dương thoả mãn \(a\sqrt{32\left(b^2+c^2\right)}+\left(b+c\right)^2=12\)

Chứng minh : \(\frac{a^3}{b+3\sqrt{bc}}+\frac{b^3}{c+3\sqrt{ca}}+\frac{c^3}{a+3\sqrt{ca}}\ge\frac{3}{4}\)

- Giải phương trình sau: \(2x^3-x^2+\sqrt[3]{2x^3-3x+1}=3x+1+\sqrt[3]{x^2+2}\)

Answer 1

[Hoàng Thị Ánh Phương]

Ta có : \(a\sqrt{32\left(b^2+c^2\right)}=2.2a\sqrt{2\left(b^2+c^2\right)}\le4a^2+2\left(b^2+c^2\right)\)

\(\left(b+c\right)^2\le2\left(b^2+c^2\right)\)

\(\Rightarrow12\le4\left(a^2+b^2+c^2\right)\Rightarrow a^2+b^2+c^2\ge3\)

Ngoài ra \(a\sqrt{\left(16+16\right)\left(b^2+b^2\right)}\ge a\left(4a+4b\right)\)

\(\left(b+c\right)^2\ge4bc\)

\(\Rightarrow ab+bc+ac\le3\)

\(VT=\frac{a^4}{ab+3a\sqrt{bc}}+\frac{b^4}{bc+3b\sqrt{ca}}+\frac{c^4}{ac+3c\sqrt{ba}}\)

\(\ge\frac{a^4}{ab+\frac{3}{2}\left(a^2+bc\right)}+\frac{b^4}{bc+\frac{3}{2}\left(b^2+ac\right)}+\frac{c^4}{ac+\frac{3}{2}\left(c^2+ab\right)}\)

\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{\frac{3}{2}\left(a^2+b^2+c^2\right)+\frac{5}{2}\left(ab+bc+ac\right)}\)

\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{\frac{3}{2}\left(a^2+b^2+c^2\right)+\frac{15}{2}}\)

Xét VT \(\ge\frac{3}{4}\)

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2\ge\frac{9}{8}\left(a^2+b^2+c^2\right)+\frac{45}{8}\)

\(\Leftrightarrow\left(a^2+b^2+c^2-3\right)+\left(a^2+b^2+c^2+\frac{15}{8}\right)\ge0\) ( luôn đúng với \(a^2+b^2+c^2\ge3\) )

\(\Rightarrowđpcm\)

Dấu " = " xảy ra khi a = b = c = 1