Question

Cho a,b,c là các số thực dương thỏa mãn a + b + c = 2. CMR

\(\frac{a}{4a+3bc}+\frac{b}{4b+3ac}+\frac{c}{4c+3ab}\) ≤ \(\frac{1}{2}\)

Answer 1

\(\Leftrightarrow\frac{4a}{4a+3bc}+\frac{4b}{4b+3ac}+\frac{4c}{4c+3ab}\le2\)

\(\Leftrightarrow\frac{bc}{4a+3bc}+\frac{ac}{4b+3ac}+\frac{ab}{4c+3ab}\ge\frac{1}{3}\)

Thật vậy, ta có:

\(VT=\frac{b^2c^2}{4abc+3b^2c^2}+\frac{a^2c^2}{4abc+3a^2c^2}+\frac{a^2b^2}{4abc+3a^2b^2}\)

\(VT\ge\frac{\left(ab+bc+ca\right)^2}{3\left(a^2b^2+b^2c^2+c^2a^2\right)+12abc}=\frac{a^2b^2+b^2c^2+c^2a^2+2\left(a+b+c\right)abc}{3\left(a^2b^2+b^2c^2+c^2a^2+4abc\right)}\)

\(VT\ge\frac{a^2b^2+b^2c^2+c^2a^2+4abc}{3\left(a^2b^2+b^2c^2+c^2a^2+4abc\right)}=\frac{1}{3}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=\frac{2}{3}\)