Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{x^2+z^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\)
\(\Rightarrow VT=\dfrac{y^2+z^2-x^2}{2x}+\dfrac{x^2+z^2-y^2}{2y}+\dfrac{x^2+y^2-z^2}{2z}\)
\(VT\ge\dfrac{\left(y+z\right)^2}{4x}+\dfrac{\left(x+z\right)^2}{4y}+\dfrac{\left(x+y\right)^2}{4z}-\dfrac{1}{2}\left(x+y+z\right)\)
\(VT\ge\dfrac{\left(2x+2y+2z\right)^2}{4\left(x+y+z\right)}-\dfrac{1}{2}\left(x+y+z\right)=\dfrac{1}{2}\left(x+y+z\right)\)
\(VT\ge\dfrac{1}{2}\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)\)
\(VT\ge\dfrac{1}{2}\left(\sqrt{\dfrac{1}{2}\left(a+b\right)^2}+\sqrt{\dfrac{1}{2}\left(b+c\right)^2}+\sqrt{\dfrac{1}{2}\left(c+a\right)^2}\right)\)
\(VT\ge\dfrac{a+b+c}{\sqrt{2}}\) (đpcm)
