Question

[Yim Yim]

Cho a+b+c>0 t/m:

\(\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{c^2+a^2}=\sqrt{2017}\)

Chứng minh rằng ;

\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{1}{2}\sqrt{\frac{2017}{2}}\)

Answer 1

[Thắng Nguyễn]

Đặt \(\hept{\begin{cases}\sqrt{a^2+b^2}=z\\\sqrt{b^2+c^2}=x\\\sqrt{c^2+a^2}=y\end{cases}}\Rightarrow\hept{\begin{cases}a=\frac{y^2+z^2-x^2}{2}\\b=\frac{x^2+z^2-y^2}{2}\\c=\frac{x^2+y^2-z^2}{2}\end{cases}}\)\(\forall\hept{\begin{cases}x,y,z>0\\x+y+z=\sqrt{2017}\end{cases}}\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(b+c\le\sqrt{2\left(b^2+c^2\right)}=2x\Rightarrow\frac{a^2}{b+c}\ge\frac{y^2+z^2-x^2}{2\sqrt{2}x}\)

Tương tự cho 2 BĐT còn lại rồi cộng theo vế:

\(2\sqrt{2}\cdot VT\ge\frac{y^2+z^2-x^2}{x}+\frac{y^2+x^2-z^2}{z}+\frac{x^2+z^2-y^2}{y}\)

\(=\frac{y^2}{x}+\frac{z^2}{x}+\frac{y^2}{z}+\frac{x^2}{z}+\frac{x^2}{y}+\frac{z^2}{y}-\left(x+y+z\right)\)

\(\ge\frac{\left(2\left(x+y+z\right)\right)^2}{2\left(x+y+z\right)}-\sqrt{2017}=\sqrt{2017}\)

\(\Rightarrow2\sqrt{2}\cdot VT\ge\sqrt{2017}\Rightarrow VT\ge\frac{\sqrt{2017}}{2\sqrt{2}}=VP\)