Chắc là a;b;c dương
Đặt \(\left(a;b;c\right)=\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\) và vế trái là P
\(P=\frac{x}{\sqrt{z\left(3x+y\right)}}+\frac{y}{\sqrt{x\left(3y+z\right)}}+\frac{z}{\sqrt{y\left(3z+x\right)}}=\frac{x^2}{x\sqrt{3xz+yz}}+\frac{y^2}{y\sqrt{3xy+xz}}+\frac{z^2}{z\sqrt{3yz+xy}}\)
\(P\ge\frac{\left(x+y+z\right)^2}{x\sqrt{3xz+yz}+y\sqrt{3xy+xz}+z\sqrt{3yz+xy}}=\frac{\left(x+y+z\right)^2}{Q}\)
\(Q=\sqrt{x\left(3x^2z+xyz\right)}+\sqrt{y\left(3xy^2+xyz\right)}+\sqrt{z\left(3yz^2+xyz\right)}\)
\(\Rightarrow Q^2\le3\left(x+y+z\right)\left(xy^2+yz^2+zx^2+xyz\right)\)
Không mất tính tổng quát, giả sử \(x=mid\left\{x;y;z\right\}\)
\(\Rightarrow\left(x-y\right)\left(x-z\right)\le0\Rightarrow x^2+yz\le xy+xz\)
\(\Rightarrow zx^2+yz^2\le xyz+xz^2\Rightarrow xy^2+yz^2+zx^2+xyz\le xy^2+2xyz+xz^2\)
\(\Rightarrow xy^2+yz^2+zx^2+xyz\le x\left(y+z\right)^2=\frac{1}{2}.2x\left(y+z\right)\left(y+z\right)\le\frac{4}{27}\left(x+y+z\right)^3\)
\(\Rightarrow Q^2\le3\left(x+y+z\right).\frac{4}{27}\left(x+y+z\right)^3=\frac{4}{9}\left(x+y+z\right)^4\)
\(\Rightarrow Q\le\frac{2}{3}\left(x+y+z\right)^2\)
\(\Rightarrow P\ge\frac{\left(x+y+z\right)^2}{\frac{2}{3}\left(x+y+z\right)^2}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
