Question

Cho 3 số dương a;b;c thỏa mãn: \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=1\)

CMR: \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{1}{2\sqrt{2}}\)

Answer 1

Đặt \(\hept{\begin{cases}\sqrt{a^2+b^2}=x\\\sqrt{b^2+c^2}=y\\\sqrt{c^2+a^2}=z\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x,y,z>0\\x+y+z=1\end{cases}}\)

Và \(\hept{\begin{cases}a^2=\frac{x^2+z^2-y^2}{2}\\b^2=\frac{x^2+y^2-z^2}{2}\\c^2=\frac{y^2+z^2-x^2}{2}\end{cases}}\) và \(\hept{\begin{cases}b+c\le\sqrt{2\left(b^2+c^2\right)}=\sqrt{2}y\\a+b\le\sqrt{2}x\\c+a\le\sqrt{2}z\end{cases}}\)

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

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

\(=\frac{1}{2\sqrt{2}}\left(x+y+z\right)=\frac{1}{2\sqrt{2}}\)