Đặt vế trái BĐT cần chứng minh là P

Ta có:

\(P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}+\dfrac{b^2}{\sqrt{2\left(a^2+c^2\right)}}+\dfrac{c^2}{\sqrt{2\left(a^2+b^2\right)}}\)

Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\Rightarrow x+y+z=\sqrt{2011}\)

Đồng thời: \(\left\{{}\begin{matrix}y^2+z^2-x^2=2a^2\\z^2+x^2-y^2=2b^2\\x^2+y^2-z^2=2c^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{z^2+x^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\)

\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{y^2+z^2-x^2}{x}+\dfrac{z^2+x^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}\right)\)

\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{y^2+z^2}{x}+\dfrac{z^2+x^2}{y}+\dfrac{x^2+y^2}{z}-\left(x+y+z\right)\right)\)

\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{\left(y+z\right)^2}{2x}+\dfrac{\left(z+x\right)^2}{2y}+\dfrac{\left(x+y\right)^2}{2z}-\left(x+y+z\right)\right)\)

\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{\left(y+z+z+x+x+y\right)^2}{2x+2y+2z}-\left(x+y+z\right)\right)=\dfrac{1}{2\sqrt{2}}\left(x+y+z\right)=\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\sqrt{\dfrac{2011}{2}}\)

\(b+c\le\sqrt{2\left(b^2+c^2\right)}\Rightarrow\dfrac{a^2}{b+c}\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}=\dfrac{1}{\sqrt{2}}.\dfrac{a^2}{\sqrt{b^2+c^2}}\)

Sau đó làm tiếp như bài đó là được

Đặt \(\left\{{}\begin{matrix}\sqrt{a^2+b^2}=x\\\sqrt{b^2+c^2}=y\\\sqrt{c^2+a^2}=z\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{x^2+z^2-y^2}{2}\\b^2=\dfrac{x^2+y^2-z^2}{2}\\c^2=\dfrac{y^2+z^2-x^2}{2}\\x+y+z=\sqrt{2011}\end{matrix}\right.\)

Và \(\left\{{}\begin{matrix}b+c\le\sqrt{2\left(b^2+c^2\right)}=\sqrt{2}y\\a+b\le\sqrt{2}x\\c+a\le\sqrt{2}z\end{matrix}\right.\)

\(VT=\dfrac{1}{2\sqrt{2}}\left(\dfrac{x^2+z^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}+\dfrac{y^2+z^2-x^2}{x}\right)\)

\(\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{2\left(x+y+z\right)^2}{\left(x+y+z\right)}-\left(x+y+z\right)\right)\)

\(=\dfrac{1}{2\sqrt{2}}\left(x+y+z\right)=\dfrac{\sqrt{2011}}{2\sqrt{2}}=VP\)

Áp dụng BĐT BSC:

\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}\)

\(=\dfrac{a+b+c}{2}\)

\(\ge\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\dfrac{1}{2}\)

Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{3}\)